b^2+15=-8b

Solution for b^2+15=-8b equation:

b^2+15=-8b

We move all terms to the left:

b^2+15-(-8b)=0

We get rid of parentheses

b^2+8b+15=0

a = 1; b = 8; c = +15;
Δ = b2-4ac
Δ = 82-4·1·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*1}=\frac{-10}{2}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*1}=\frac{-6}{2}
=-3
$