b^2+3b=60

Solution for b^2+3b=60 equation:

b^2+3b=60

We move all terms to the left:

b^2+3b-(60)=0

a = 1; b = 3; c = -60;
Δ = b2-4ac
Δ = 32-4·1·(-60)
Δ = 249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{249}}{2*1}=\frac{-3-\sqrt{249}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{249}}{2*1}=\frac{-3+\sqrt{249}}{2}
$