Question

Solve the following equation:
$\frac{x^2-12x-20}{3x-5}=\frac{x^2+8x+12}{2x+3}$

Answer 1

$\frac{x^2-12x+20}{3x-5}=\frac{x^2-8x+12}{2x+3}$

$\Rightarrow \frac{x^2-10x-2x-20}{3x-5}=\frac{x^2-6x-2x+12}{2x+3}$

$\Rightarrow \frac{x(x-10)-2(x-10)}{3x-5}=\frac{x(x-6)-2(x-6)}{2x+3}$

$\Rightarrow \frac{(x-10)(x-2)}{3x-5}=\frac{(x-6)(x-2)}{2x+3}$

$\Rightarrow \frac{x-10}{3x-5}=\frac{x-6}{2x+3}$

$\Rightarrow (x-10)(2x+3)=(x-6)(3x-5)$

$\Rightarrow 2x^2+3x-20x-30=3x^2-5x-18x+30$

$\Rightarrow 2x^2-17x-30=3x^2-23x+30$

$\Rightarrow 3x^2-23x+30-2x^2+17x+30=0$

$\Rightarrow x^2-6x+60=0$

$\Rightarrow x^2-2\times 3x+3^2-3^2+60=0$

$\Rightarrow {(x-3)}^2-9+60=0$

$\Rightarrow {(x-3)}^2=-51$

$\Rightarrow x=3\pm \sqrt{-51}$

$\therefore x=3\pm i\sqrt{51}$