2x^2+28x-12=x^2+20x+36

Solution for 2x^2+28x-12=x^2+20x+36 equation:

2x^2+28x-12=x^2+20x+36

We move all terms to the left:

2x^2+28x-12-(x^2+20x+36)=0

We get rid of parentheses

2x^2-x^2+28x-20x-36-12=0

We add all the numbers together, and all the variables

x^2+8x-48=0

a = 1; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·1·(-48)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*1}=\frac{-24}{2}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*1}=\frac{8}{2}
=4
$