(h-4)(h+4)h=120

Solution for (h-4)(h+4)h=120 equation:

(h-4)(h+4)h=120

We move all terms to the left:

(h-4)(h+4)h-(120)=0

We use the square of the difference formula

h^2-16-120=0

We add all the numbers together, and all the variables

h^2-136=0

a = 1; b = 0; c = -136;
Δ = b2-4ac
Δ = 02-4·1·(-136)
Δ = 544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{34}}{2*1}=\frac{0-4\sqrt{34}}{2}
=-\frac{4\sqrt{34}}{2}
=-2\sqrt{34}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{34}}{2*1}=\frac{0+4\sqrt{34}}{2}
=\frac{4\sqrt{34}}{2}
=2\sqrt{34}
$