(2h-5)(5h-1)=0

Solution for (2h-5)(5h-1)=0 equation:

(2h-5)(5h-1)=0

We multiply parentheses ..

(+10h^2-2h-25h+5)=0

We get rid of parentheses

10h^2-2h-25h+5=0

We add all the numbers together, and all the variables

10h^2-27h+5=0

a = 10; b = -27; c = +5;
Δ = b2-4ac
Δ = -272-4·10·5
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-23}{2*10}=\frac{4}{20}
=1/5
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+23}{2*10}=\frac{50}{20}
=2+1/2
$