2h+4h+2h+h2=20

Solution for 2h+4h+2h+h2=20 equation:

2h+4h+2h+h2=20

We move all terms to the left:

2h+4h+2h+h2-(20)=0

We add all the numbers together, and all the variables

h^2+8h-20=0

a = 1; b = 8; c = -20;
Δ = b2-4ac
Δ = 82-4·1·(-20)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*1}=\frac{-20}{2}
=-10
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*1}=\frac{4}{2}
=2
$