H=20-0,15t^2

Solution for H=20-0,15t^2 equation:

=20-0.15H^2

We move all terms to the left:

-(20-0.15H^2)=0

We get rid of parentheses

0.15H^2-20=0

a = 0.15; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·0.15·(-20)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*0.15}=\frac{0-2\sqrt{3}}{0.3}
=-\frac{2\sqrt{3}}{0.3}
=-\frac{\sqrt{3}}{0.15}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*0.15}=\frac{0+2\sqrt{3}}{0.3}
=\frac{2\sqrt{3}}{0.3}
=\frac{\sqrt{3}}{0.15}
$