b^2+3b=18

Solution for b^2+3b=18 equation:

b^2+3b=18

We move all terms to the left:

b^2+3b-(18)=0

a = 1; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*1}=\frac{-12}{2}
=-6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*1}=\frac{6}{2}
=3
$