x^2-20x-42=10

Solution for x^2-20x-42=10 equation:

x^2-20x-42=10

We move all terms to the left:

x^2-20x-42-(10)=0

We add all the numbers together, and all the variables

x^2-20x-52=0

a = 1; b = -20; c = -52;
Δ = b2-4ac
Δ = -202-4·1·(-52)
Δ = 608
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:

$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{38}}{2*1}=\frac{20-4\sqrt{38}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{38}}{2*1}=\frac{20+4\sqrt{38}}{2}
$