2x^2-8x+32=25

Solution for 2x^2-8x+32=25 equation:

2x^2-8x+32=25

We move all terms to the left:

2x^2-8x+32-(25)=0

We add all the numbers together, and all the variables

2x^2-8x+7=0

a = 2; b = -8; c = +7;
Δ = b2-4ac
Δ = -82-4·2·7
Δ = 8
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:

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