10=(5+2h)(4+2h)-20

Solution for 10=(5+2h)(4+2h)-20 equation:

10=(5+2h)(4+2h)-20

We move all terms to the left:

10-((5+2h)(4+2h)-20)=0

We add all the numbers together, and all the variables

-((2h+5)(2h+4)-20)+10=0

We multiply parentheses ..

-((+4h^2+8h+10h+20)-20)+10=0


We calculate terms in parentheses: -((+4h^2+8h+10h+20)-20), so:

(+4h^2+8h+10h+20)-20

We get rid of parentheses

4h^2+8h+10h+20-20

We add all the numbers together, and all the variables

4h^2+18h

Back to the equation:

-(4h^2+18h)


We get rid of parentheses

-4h^2-18h+10=0

a = -4; b = -18; c = +10;
Δ = b2-4ac
Δ = -182-4·(-4)·10
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*-4}=\frac{-4}{-8}
=1/2
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*-4}=\frac{40}{-8}
=-5
$