13=4+b2

Solution for 13=4+b2 equation:

13=4+b2

We move all terms to the left:

13-(4+b2)=0

We add all the numbers together, and all the variables

-(+b^2+4)+13=0

We get rid of parentheses

-b^2-4+13=0

We add all the numbers together, and all the variables

-1b^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:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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