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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+h^2+5h-2h-10)), so:

(+h^2+5h-2h-10)

We get rid of parentheses

h^2+5h-2h-10

We add all the numbers together, and all the variables

h^2+3h-10

Back to the equation:

-(h^2+3h-10)


We get rid of parentheses

-h^2-3h+10=0

We add all the numbers together, and all the variables

-1h^2-3h+10=0

a = -1; b = -3; c = +10;
Δ = b2-4ac
Δ = -32-4·(-1)·10
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{-(-3)-7}{2*-1}=\frac{-4}{-2}
=+2
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-1}=\frac{10}{-2}
=-5
$