4x^2-40x+100=50

Solution for 4x^2-40x+100=50 equation:

4x^2-40x+100=50

We move all terms to the left:

4x^2-40x+100-(50)=0

We add all the numbers together, and all the variables

4x^2-40x+50=0

a = 4; b = -40; c = +50;
Δ = b2-4ac
Δ = -402-4·4·50
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20\sqrt{2}}{2*4}=\frac{40-20\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20\sqrt{2}}{2*4}=\frac{40+20\sqrt{2}}{8}
$