2x^2-8x+20=x^2-4x+32

Solution for 2x^2-8x+20=x^2-4x+32 equation:

2x^2-8x+20=x^2-4x+32

We move all terms to the left:

2x^2-8x+20-(x^2-4x+32)=0

We get rid of parentheses

2x^2-x^2-8x+4x-32+20=0

We add all the numbers together, and all the variables

x^2-4x-12=0

a = 1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*1}=\frac{12}{2}
=6
$