2x^2-8x+16=20

Solution for 2x^2-8x+16=20 equation:

2x^2-8x+16=20

We move all terms to the left:

2x^2-8x+16-(20)=0

We add all the numbers together, and all the variables

2x^2-8x-4=0

a = 2; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·2·(-4)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{6}}{2*2}=\frac{8-4\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{6}}{2*2}=\frac{8+4\sqrt{6}}{4}
$