(h-9)(h-10)=0

Solution for (h-9)(h-10)=0 equation:

(h-9)(h-10)=0

We multiply parentheses ..

(+h^2-10h-9h+90)=0

We get rid of parentheses

h^2-10h-9h+90=0

We add all the numbers together, and all the variables

h^2-19h+90=0

a = 1; b = -19; c = +90;
Δ = b2-4ac
Δ = -192-4·1·90
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*1}=\frac{18}{2}
=9
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*1}=\frac{20}{2}
=10
$