(2x+5)(2x-4)-3x^2+12=0

Solution for (2x+5)(2x-4)-3x^2+12=0 equation:

(2x+5)(2x-4)-3x^2+12=0

We multiply parentheses ..

-3x^2+(+4x^2-8x+10x-20)+12=0

We get rid of parentheses

-3x^2+4x^2-8x+10x-20+12=0

We add all the numbers together, and all the variables

x^2+2x-8=0

a = 1; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*1}=\frac{4}{2}
=2
$