h=-0.15t2+5t

Solution for h=-0.15t2+5t equation:

=-0.15H^2+5H

We move all terms to the left:

-(-0.15H^2+5H)=0

We get rid of parentheses

0.15H^2-5H=0

a = 0.15; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·0.15·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*0.15}=\frac{0}{0.3}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*0.15}=\frac{10}{0.3}
=33+0.1/0.3
$