b(b+3)=20

Solution for b(b+3)=20 equation:

b(b+3)=20

We move all terms to the left:

b(b+3)-(20)=0

We multiply parentheses

b^2+3b-20=0

a = 1; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·1·(-20)
Δ = 89
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{89}}{2*1}=\frac{-3-\sqrt{89}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{89}}{2*1}=\frac{-3+\sqrt{89}}{2}
$