b^2+42=-13b

Solution for b^2+42=-13b equation:

b^2+42=-13b

We move all terms to the left:

b^2+42-(-13b)=0

We get rid of parentheses

b^2+13b+42=0

a = 1; b = 13; c = +42;
Δ = b2-4ac
Δ = 132-4·1·42
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*1}=\frac{-14}{2}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*1}=\frac{-12}{2}
=-6
$