h^2-4h=96

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

h^2-4h=96

We move all terms to the left:

h^2-4h-(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:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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