X^2-40x+8=4

Solution for X^2-40x+8=4 equation:

X^2-40X+8=4

We move all terms to the left:

X^2-40X+8-(4)=0

We add all the numbers together, and all the variables

X^2-40X+4=0

a = 1; b = -40; c = +4;
Δ = b2-4ac
Δ = -402-4·1·4
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-12\sqrt{11}}{2*1}=\frac{40-12\sqrt{11}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+12\sqrt{11}}{2*1}=\frac{40+12\sqrt{11}}{2}
$