x^2+2=10x+20

Solution for x^2+2=10x+20 equation:

x^2+2=10x+20

We move all terms to the left:

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

We get rid of parentheses

x^2-10x-20+2=0

We add all the numbers together, and all the variables

x^2-10x-18=0

a = 1; b = -10; c = -18;
Δ = b2-4ac
Δ = -102-4·1·(-18)
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{43}}{2*1}=\frac{10-2\sqrt{43}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{43}}{2*1}=\frac{10+2\sqrt{43}}{2}
$