H=500+300t-9.8t^2

Solution for H=500+300t-9.8t^2 equation:

=500+300H-9.8H^2

We move all terms to the left:

-(500+300H-9.8H^2)=0

We get rid of parentheses

9.8H^2-300H-500=0

a = 9.8; b = -300; c = -500;
Δ = b2-4ac
Δ = -3002-4·9.8·(-500)
Δ = 109600
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:

$\sqrt{\Delta}=\sqrt{109600}=\sqrt{400*274}=\sqrt{400}*\sqrt{274}=20\sqrt{274}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-300)-20\sqrt{274}}{2*9.8}=\frac{300-20\sqrt{274}}{19.6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-300)+20\sqrt{274}}{2*9.8}=\frac{300+20\sqrt{274}}{19.6}
$