b^2-4b=96

Solution for b^2-4b=96 equation:

b^2-4b=96

We move all terms to the left:

b^2-4b-(96)=0

a = 1; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*1}=\frac{-16}{2}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*1}=\frac{24}{2}
=12
$