x^2+2=2x+38

Solution for x^2+2=2x+38 equation:

x^2+2=2x+38

We move all terms to the left:

x^2+2-(2x+38)=0

We get rid of parentheses

x^2-2x-38+2=0

We add all the numbers together, and all the variables

x^2-2x-36=0

a = 1; b = -2; c = -36;
Δ = b2-4ac
Δ = -22-4·1·(-36)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{37}}{2*1}=\frac{2-2\sqrt{37}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{37}}{2*1}=\frac{2+2\sqrt{37}}{2}
$