2x^2-8x+16=400

Solution for 2x^2-8x+16=400 equation:

2x^2-8x+16=400

We move all terms to the left:

2x^2-8x+16-(400)=0

We add all the numbers together, and all the variables

2x^2-8x-384=0

a = 2; b = -8; c = -384;
Δ = b2-4ac
Δ = -82-4·2·(-384)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-56}{2*2}=\frac{-48}{4}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+56}{2*2}=\frac{64}{4}
=16
$