b^2+20b=96^2

Solution for b^2+20b=96^2 equation:

b^2+20b=96^2

We move all terms to the left:

b^2+20b-(96^2)=0

We add all the numbers together, and all the variables

b^2+20b-9216=0

a = 1; b = 20; c = -9216;
Δ = b2-4ac
Δ = 202-4·1·(-9216)
Δ = 37264
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{37264}=\sqrt{16*2329}=\sqrt{16}*\sqrt{2329}=4\sqrt{2329}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{2329}}{2*1}=\frac{-20-4\sqrt{2329}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{2329}}{2*1}=\frac{-20+4\sqrt{2329}}{2}
$