0.1h(h+20)=3

Solution for 0.1h(h+20)=3 equation:

0.1h(h+20)=3

We move all terms to the left:

0.1h(h+20)-(3)=0

We multiply parentheses

0h^2+0h-3=0

We add all the numbers together, and all the variables

h^2+h-3=0

a = 1; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·1·(-3)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{13}}{2*1}=\frac{-1-\sqrt{13}}{2}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*1}=\frac{-1+\sqrt{13}}{2}
$