2x^2-4x+8=10

Solution for 2x^2-4x+8=10 equation:

2x^2-4x+8=10

We move all terms to the left:

2x^2-4x+8-(10)=0

We add all the numbers together, and all the variables

2x^2-4x-2=0

a = 2; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·2·(-2)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{2}}{2*2}=\frac{4-4\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{2}}{2*2}=\frac{4+4\sqrt{2}}{4}
$