2x^2-8x+16=18

Solution for 2x^2-8x+16=18 equation:

2x^2-8x+16=18

We move all terms to the left:

2x^2-8x+16-(18)=0

We add all the numbers together, and all the variables

2x^2-8x-2=0

a = 2; b = -8; c = -2;
Δ = b2-4ac
Δ = -82-4·2·(-2)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{5}}{2*2}=\frac{8-4\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{5}}{2*2}=\frac{8+4\sqrt{5}}{4}
$