(h-4)(3h-5)=0

Solution for (h-4)(3h-5)=0 equation:

(h-4)(3h-5)=0

We multiply parentheses ..

(+3h^2-5h-12h+20)=0

We get rid of parentheses

3h^2-5h-12h+20=0

We add all the numbers together, and all the variables

3h^2-17h+20=0

a = 3; b = -17; c = +20;
Δ = b2-4ac
Δ = -172-4·3·20
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-7}{2*3}=\frac{10}{6}
=1+2/3
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+7}{2*3}=\frac{24}{6}
=4
$