b^2+15b=-50

Solution for b^2+15b=-50 equation:

b^2+15b=-50

We move all terms to the left:

b^2+15b-(-50)=0

We add all the numbers together, and all the variables

b^2+15b+50=0

a = 1; b = 15; c = +50;
Δ = b2-4ac
Δ = 152-4·1·50
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*1}=\frac{-20}{2}
=-10
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*1}=\frac{-10}{2}
=-5
$