b^2+20b-99=-3

Solution for b^2+20b-99=-3 equation:

b^2+20b-99=-3

We move all terms to the left:

b^2+20b-99-(-3)=0

We add all the numbers together, and all the variables

b^2+20b-96=0

a = 1; b = 20; c = -96;
Δ = b2-4ac
Δ = 202-4·1·(-96)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*1}=\frac{-48}{2}
=-24
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*1}=\frac{8}{2}
=4
$