Question

Solve for x : $ab{x}^2+(b^2-ac)x-bc=0$

Answer 1

$ab{x}^2+(b^2-ac)x-bc=0$

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(b^2-ac)\pm \sqrt{{(b^2-ac)}^2-4ab(-bc)}}{2ab}$

$x=\frac{-b^2+ac\pm \sqrt{a^2{c}^2+2a{b}^{2}c+b^4}}{2ab}$