8x+12x^2=9x^2-40x

Solution for 8x+12x^2=9x^2-40x equation:

8x+12x^2=9x^2-40x

We move all terms to the left:

8x+12x^2-(9x^2-40x)=0

We get rid of parentheses

12x^2-9x^2+8x+40x=0

We add all the numbers together, and all the variables

3x^2+48x=0

a = 3; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·3·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*3}=\frac{-96}{6}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*3}=\frac{0}{6}
=0
$