2x^2-8x+8=40

Solution for 2x^2-8x+8=40 equation:

2x^2-8x+8=40

We move all terms to the left:

2x^2-8x+8-(40)=0

We add all the numbers together, and all the variables

2x^2-8x-32=0

a = 2; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·2·(-32)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{5}}{2*2}=\frac{8-8\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{5}}{2*2}=\frac{8+8\sqrt{5}}{4}
$