3x(x-3)=8x^2-12x

Solution for 3x(x-3)=8x^2-12x equation:

3x(x-3)=8x^2-12x

We move all terms to the left:

3x(x-3)-(8x^2-12x)=0

We multiply parentheses

3x^2-9x-(8x^2-12x)=0

We get rid of parentheses

3x^2-8x^2-9x+12x=0

We add all the numbers together, and all the variables

-5x^2+3x=0

a = -5; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-5)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-5}=\frac{-6}{-10}
=3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-5}=\frac{0}{-10}
=0
$