(h+4)(h-3)=60

Solution for (h+4)(h-3)=60 equation:

(h+4)(h-3)=60

We move all terms to the left:

(h+4)(h-3)-(60)=0

We multiply parentheses ..

(+h^2-3h+4h-12)-60=0

We get rid of parentheses

h^2-3h+4h-12-60=0

We add all the numbers together, and all the variables

h^2+h-72=0

a = 1; b = 1; c = -72;
Δ = b2-4ac
Δ = 12-4·1·(-72)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*1}=\frac{-18}{2}
=-9
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*1}=\frac{16}{2}
=8
$