(h-9)(h-5)=0

Solution for (h-9)(h-5)=0 equation:

(h-9)(h-5)=0

We multiply parentheses ..

(+h^2-5h-9h+45)=0

We get rid of parentheses

h^2-5h-9h+45=0

We add all the numbers together, and all the variables

h^2-14h+45=0

a = 1; b = -14; c = +45;
Δ = b2-4ac
Δ = -142-4·1·45
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*1}=\frac{10}{2}
=5
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*1}=\frac{18}{2}
=9
$