\begin{align*}
x^{2} +2x -15 &= 0 \\
(x -3)(x +5) &=0 \\
\therefore x = -5 & \text{ or } x = 3
\end{align*}
\begin{align*}
p^{2} -7p -18 &= 0 \\
(p -9)(p +2) =&0 \\
\therefore p = -2 &\text{ or } p = 9
\end{align*}
\begin{align*}
9x^{2} - 6x - 8 & = 0 \\
(3x + 2)(3x - 4) & = 0 \\
3x + 2 & = 0 \\
x & = -\frac{2}{3} \\
\text{or} & \\
3x - 4 & = 0 \\
x & = \frac{4}{3} \\
\therefore x = -\frac{2}{3} & \text{ or } x = \frac{4}{3}
\end{align*}
\(5x^{2} + 21x - 54 = 0\)
\begin{align*}
5x^{2} + 21x - 54 & = 0 \\
(5x - 9)(x + 6) & = 0 \\
5x - 9 & = 0 \\
x & = \frac{9}{5} \\
\text{or} & \\
x + 6 & = 0 \\
x & = -6 \\
\therefore x = \frac{9}{5} & \text{ or } x = -6
\end{align*}
\(4 z^{2} + 12 z + 8 = 0\)
\begin{align*}
4 z^{2} + 12 z + 8 &= 0\\
z^{2} + 3 z + 2 &= 0 \\
\left(z + 1\right) \left(z + 2\right)&= 0 \\
z = -2 &\text{ or } z = -1
\end{align*}
\(- b^{2} + 7 b - 12 = 0\)
\begin{align*}
- b^{2} + 7 b - 12 &= 0 \\
b^{2} - 7 b + 12 &= 0 \\
\left(b - 4\right) \left(b - 3\right) &= 0 \\
b = 3 &\text{ or } b = 4
\end{align*}
\(- 3 a^{2} + 27 a - 54 = 0\)
\begin{align*}
- 3 a^{2} + 27 a - 54 &= 0 \\
a^{2} - 9 a + 18 &= 0 \\
\left(a - 6\right) \left(a - 3\right) &= 0 \\
a = 3 &\text{ or } a = 6.
\end{align*}
\begin{align*}
4y^{2} - 9 & = 0 \\
(2y - 3)(2y + 3) & = 0 \\
2y - 3 & = 0 \\
y & = \frac{3}{2} \\
\text{or} & \\
2y + 3 & = 0 \\
y & = -\frac{3}{2} \\
\therefore y = \frac{3}{2} & \text{ or } y = -\frac{3}{2}
\end{align*}
\begin{align*}
4x^{2} + 16x - 9 & = 0 \\
(2x - 1)(2x + 9) & = 0 \\
2x - 1 & = 0 \\
x & = \frac{1}{2} \\
\text{or} & \\
2x + 9 & = 0 \\
y & = -\frac{9}{2} \\
\therefore x = \frac{1}{2} & \text{ or } x = -\frac{9}{2}
\end{align*}
\begin{align*}
4x^{2} - 12x & = -9 \\
4x^{2} - 12x + 9 & = 0 \\
(2x - 3)(2x - 3) & = 0 \\
2x - 3 & = 0 \\
x & = \frac{3}{2}
\end{align*}
\begin{align*}
20m + 25m^{2} & = 0 \\
5m(4 + 5m) & = 0 \\
5m & = 0 \\
m & = 0 \\
\text{ or } & \\
4 + 5m & = 0 \\
m & = -\frac{4}{5} \\
\therefore m = 0 & \text{ or } m = -\frac{4}{5}
\end{align*}
\begin{align*}
2x^{2} - 5x - 12 & = 0 \\
(2x + 3)(x - 4) & = 0 \\
2x + 3 & = 0 \\
x & = -\frac{3}{2} \\
\text{or} & \\
x - 4 & = 0 \\
x & = 4 \\
\therefore x = -\frac{3}{2} & \text{ or } x = 4
\end{align*}
\(-75x^{2} + 290x = 240\)
\begin{align*}
-75x^{2} + 290x & = 240 \\
-75x^{2} + 290x - 240 & = 0 \\
-15x^{2} + 58x - 48 & = 0 \\
(5x - 6)(3x - 8) & = 0 \\
5x - 6 & = 0 \\
x & = \frac{6}{5} \\
\text{or} & \\
3x - 8 & = 0 \\
x & = \frac{8}{3} \\
\therefore x = \frac{6}{5} & \text{ or } x = \frac{8}{3}
\end{align*}
\(2x = \frac{1}{3}x^{2} - 3x + 14\frac{2}{3}\)
\begin{align*}
2x & = \frac{1}{3}x^{2} - 3x + 14\frac{2}{3} \\
6x & = x^{2} - 9x + 44 \\
x^{2} - 15x + 44 & = 0 \\
(x - 4)(x - 11) & = 0 \\
x - 4 & = 0 \\
x & = 4 \\
\text{or} & \\
x - 11 & = 0 \\
x & = 11 \\
\therefore x = 4 & \text{ or } x = 11
\end{align*}
\begin{align*}
x^{2} - 4x & = -4 \\
x^{2} - 4x + 4 & = 0 \\
(x - 2)(x - 2) & = 0 \\
x - 2 & = 0 \\
x & = 2
\end{align*}
\(-x^{2} + 4x - 6 = 4x^{2} - 14x + 3\)
\begin{align*}
-x^{2} + 4x - 6 & = 4x^{2} - 14x + 3 \\
5x^{2} - 18x + 9 & = 0 \\
(5x - 3)(x - 3) & = 0 \\
5x - 3 & = 0 \\
x & = \frac{3}{5} \\
\text{or} & \\
x - 3 & = 0 \\
x & = 3 \\
\therefore x = \frac{3}{5} & \text{ or } x = 3
\end{align*}
\begin{align*}
t^{2} & = 3t \\
t^{2} - 3t & = 0 \\
t(t - 3) & = 0 \\
t & = 0 \\
\text{or} & \\
t - 3 & = 0 \\
t & = 3 \\
\therefore t = 0 & \text{ or } t = 3
\end{align*}
\begin{align*}
x^{2} - 10x & = -25 \\
x^{2} - 10x + 25 & = 0 \\
(x - 5)(x - 5) & = 0 \\
x - 5 & = 0 \\
x & = 5
\end{align*}
\begin{align*}
x^{2} & = 18 \\
\therefore x = \sqrt{18} & \text{ or } x = -\sqrt{18}
\end{align*}
\begin{align*}
p^{2} - 6p & = 7 \\
p^{2} - 6p - 7 & = 0 \\
(p - 7)(p + 1) & = 0 \\
p - 7 & = 0 \\
p & = 7 \\
\text{or} & \\
p + 1 & = 0 \\
p & = -1 \\
\therefore p = 7 & \text{ or } p = -1
\end{align*}
\(4x^{2} - 17x - 77 = 0\)
\begin{align*}
4x^{2} - 17x - 77 & = 0 \\
(4x + 11)(x - 7) & = 0 \\
4x + 11 & = 0 \\
x & = -\frac{11}{4} \\
\text{or} & \\
x - 7 & = 0 \\
x & = 7 \\
\therefore x = -\frac{11}{4} & \text{ or } x = 7
\end{align*}
\begin{align*}
14x^{2} + 5x & = 6 \\
14x^{2} + 5x - 6 & = 0 \\
(7x + 6)(2x - 1) & = 0 \\
7x + 6 & = 0 \\
x & = -\frac{6}{7} \\
\text{or} & \\
2x - 1 & = 0 \\
x & = \frac{1}{2} \\
\therefore x = -\frac{6}{7} & \text{ or } x = \frac{1}{2}
\end{align*}
\begin{align*}
2x^{2} - 2x & = 12 \\
x^{2} - x - 6 & = 0 \\
(x - 3)(x + 2) & = 0 \\
x - 3 & = 0 \\
x & = 3 \\
\text{or} & \\
x + 2 & = 0 \\
x & = -2 \\
\therefore x = 3 & \text{ or } x = -2
\end{align*}
\((2a - 3)^2 - 16 = 0\)
\begin{align*}
(2a - 3)^2 - 16 &= 0 \\
(2a - 3 + 4)(2a - 3 - 4) &= 0 \\
(2a + 1)(2a - 7)&= 0 \\
\therefore a = -\frac{1}{2} &\text{ or } a = \text{3,5}
\end{align*}
\((x-6)^2 - 24 = 1\)
\begin{align*}
(x-6)^2 - 24 &= 1 \\
(x-6)^2 - 25 &= 0 \\
(x- 6 - 5)(x- 6 + 5) &= 0\\
(x-11)(x-1) &= 0\\
\therefore x = 11 &\text{ or } x = 1
\end{align*}
\(3y = \dfrac{54}{2y}\)
Note \(y \neq 0\)
\begin{align*}
3y &= \frac{54}{2y} \\
3y^2 &= 27 \\
y^2 &= 9 \\
y^2 - 9 &= 0 \\
(y-3)(y+3) &= 0 \\
\therefore y = 3 &\text{ or } y = -3
\end{align*}
\(\dfrac{10z}{3} = 1 - \dfrac{1}{3z}\)
Note \(z \neq 0\)
\begin{align*}
\frac{10z}{3} &= 1 - \frac{1}{3z} \\
10z^2 &= 3z - 1 \\
10z^2 - 3z + 1 &= 0 \\
(5z + 1)(2z - 1) &= 0 \\
\therefore z = -\frac{1}{5} &\text{ or } z = \frac{1}{2}
\end{align*}
\(x + 2 = \dfrac{18}{x} -1\)
Note \(x \neq 0\)
\begin{align*}
x + 2 &= \frac{18}{x} -1 \\
x^2 + 2x &= 18 - x \\
x^2 + 3x - 18 &= 0 \\
(x-3)(x+6) &= 0 \\
\therefore x = 3 &\text{ or } x = -6
\end{align*}
\(y - 3 = \dfrac{5}{4} - \dfrac{1}{y}\)
Note \(y \neq 0\)
\begin{align*}
y - 3 &= \frac{5}{4} - \frac{1}{y} \\
4y^2 - 12y &= 5y - 4 \\
4y^2 -17y + 4 &= 0\\
(4y-1)(y-4)&= 0\\
\therefore y = \frac{1}{4} &\text{ or } y = 4
\end{align*}
\(\dfrac{1}{2}(b - 1) = \dfrac{1}{3}\left(\dfrac{2}{b} + 4\right)\)
Note \(b \neq 0\)
\begin{align*}
\frac{1}{2}(b - 1) &= \frac{1}{3}\left(\frac{2}{b} + 4\right) \\
3(b - 1) &= 2\left(\frac{2}{b} + 4\right) \\
3b - 3 &= \frac{4}{b} + 8 \\
3b^2 - 3b &= 4 + 8b \\
3b^2 - 11b - 4 &= 0 \\
(3b + 1)(b - 4) &= \\
\therefore b = -\frac{1}{3}b &\text{ or } b = 4
\end{align*}
\(3(y + 1) = \dfrac{4}{y} + 2\)
Note \(y \neq 0\)
\begin{align*}
3(y + 1) &= \frac{4}{y} + 2 \\
3y + 3 &= \frac{4}{y} + 2 \\
3y^2 + 3y &= 4 + 2y \\
3y^2 + y - 4 &= 0 \\
(3y + 4)(y-1) &= 0 \\
\therefore y = -\frac{4}{3} &\text{ or } y = 1
\end{align*}
\((x+1)^2 - 2(x+1) - 15 = 0\)
\begin{align*}
(x+1)^2 - 2(x+1) - 15 &= 0 \\
((x+1) - 5)((x+1) + 3) &= 0\\
(x-4)(x+4) &= 0 \\
\therefore x = 4 &\text{ or } x = -4
\end{align*}
\(z^4 - 1 = 0\)
\begin{align*}
z^4 - 1 &= 0 \\
(z^2 - 1)(z^2 + 1) &= 0 \\
(z- 1)(z+1)(z^2 + 1) &= 0 \\
\therefore z = 1 &\text{ or } z = -1
\end{align*}
Note that \(z^{2} + 1\) has no real solutions.
\(b^4 - 13b^2 + 36 = 0\)
\begin{align*}
b^4 - 13b^2 + 36 &= 0 \\
(b^2 - 4)(b^2 - 9) &= 0 \\
(b-2)(b+2)(b-3)(b+3) &= 0 \\
\therefore b = \pm 2 &\text{ or } b = \pm 3
\end{align*}
\(\dfrac{a + 1}{3a - 4} + \dfrac{9}{2a + 5} + \dfrac{2a + 3}{2a + 5} = 0\)
\begin{align*}
\frac{a + 1}{3a - 4} + \frac{9}{2a + 5} + \frac{2a + 3}{2a + 5} & = 0 \\
\frac{(a + 1)(2a + 5) + 9(3a - 4) + (2a + 3)(3a - 4)}{(3a - 4)(2a + 5)} & = 0 \\
2a^{2} + 7a + 5 + 27a - 36 + 6a^{2} + a - 12 & = 0 \\
8a^{2} + 35a - 43 & = 0 \\
(8a + 43)(a - 1) & = 0 \\
8a + 43 & = 0 \\
a & = -\frac{43}{8} \\
\text{or} & \\
a - 1 & = 0 \\
a & = 1 \\
\therefore a = -\frac{43}{8} & \text{ or } a = 1
\end{align*}
\(\dfrac{x^2 - 2x - 3}{x+1} = 0\)
Note \(x \neq -1\)
\begin{align*}
\frac{x^2 - 2x - 3}{x+1} &= 0 \\
\frac{(x+1)(x-3)}{x+1} &= 0 \\
\therefore x &= 3
\end{align*}
\(x + 2 = \dfrac{6x -12}{x- 2}\)
Note \(x \neq 2\)
\begin{align*}
x + 2 &= \frac{6x-12}{x- 2} \\
(x+2)(x-2) &= 6x - 12 \\
x^2 - 4 &= 6x - 12\\
x^2 - 6x + 8 & = 0 \\
(x - 2)(x - 4) & = 0 \\
\therefore x &= 4
\end{align*}
\(\dfrac{3(a^2+1) +10a}{3a + 1} = 1\)
Note \(a \neq -\frac{1}{3}\)
\begin{align*}
\frac{3(a^2+1)+ 10a}{3a + 1} &= 1 \\
3(a^2+1) + 10a &= 3a + 1 \\
3a^2 + 3 + 10a - 3a - 1 &= 0 \\
3a^2 + 7a + 2 &= 0 \\
(3a + 1)(a + 2) &= 0 \\
\therefore a &= -2
\end{align*}
\(\dfrac{3}{9a^{2} - 3a + 1} - \dfrac{3a + 4}{27a^{3} + 1} = \dfrac{1}{9a^{2} - 1}\)
\begin{align*}
\frac{3}{9a^{2} - 3a + 1} - \frac{3a + 4}{27a^{3} + 1} & = \frac{1}{9a^{2} - 1} \\
\frac{3}{9a^{2} - 3a + 1} - \frac{3a + 4}{(3a + 1)(9a^{2} - 3a + 1)} & =
\frac{1}{(3a - 1)(3a + 1)} \\
\frac{3(9a^{2} - 1) - (3a - 1)(3a + 4)}{(3a + 1)(3a - 1)(9a^{2} - 3a + 1)} & =
\frac{9a^{2} - 3a + 1}{(3a - 1)(3a + 1)(9a^{2} - 3a + 1)} \\
27a^{2} - 3 - 9a^{2} - 9a + 4 & = 9a^{2} - 3a + 1 \\
9a^{2} - 6a & = 0 \\
3a(3a - 2) & = 0 \\
3a & = 0 \\
a & = 0 \\
\text{or} & \\
3a - 2 & = 0 \\
a & = \frac{2}{3} \\
\therefore a = 0 & \text{ or } a = \frac{2}{3}
\end{align*}