13=x^2+4

Solution for 13=x^2+4 equation:

13=x^2+4

We move all terms to the left:

13-(x^2+4)=0

We get rid of parentheses

-x^2-4+13=0

We add all the numbers together, and all the variables

-1x^2+9=0

a = -1; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-1)·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-1}=\frac{6}{-2}
=-3
$