H+(h-8)+h(h-8)=72

Solution for H+(h-8)+h(h-8)=72 equation:

+(H-8)+H(H-8)=72

We move all terms to the left:

+(H-8)+H(H-8)-(72)=0

We multiply parentheses

H^2+(H-8)-8H-72=0

We get rid of parentheses

H^2+H-8H-8-72=0

We add all the numbers together, and all the variables

H^2-7H-80=0

a = 1; b = -7; c = -80;
Δ = b2-4ac
Δ = -72-4·1·(-80)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3\sqrt{41}}{2*1}=\frac{7-3\sqrt{41}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3\sqrt{41}}{2*1}=\frac{7+3\sqrt{41}}{2}
$