H=-4.9t^2+96

Solution for H=-4.9t^2+96 equation:

=-4.9H^2+96

We move all terms to the left:

-(-4.9H^2+96)=0

We get rid of parentheses

4.9H^2-96=0

a = 4.9; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·4.9·(-96)
Δ = 1881.6
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{-(0)-\sqrt{1881.6}}{2*4.9}=\frac{0-\sqrt{1881.6}}{9.8}
=-\frac{\sqrt{}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1881.6}}{2*4.9}=\frac{0+\sqrt{1881.6}}{9.8}
=\frac{\sqrt{}}{9.8}
$