H=-4.9t2+100

Solution for H=-4.9t2+100 equation:

=-4.9H^2+100

We move all terms to the left:

-(-4.9H^2+100)=0

We get rid of parentheses

4.9H^2-100=0

a = 4.9; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·4.9·(-100)
Δ = 1960
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:

$\sqrt{\Delta}=\sqrt{1960}=\sqrt{4*490}=\sqrt{4}*\sqrt{490}=2\sqrt{490}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{490}}{2*4.9}=\frac{0-2\sqrt{490}}{9.8}
=-\frac{2\sqrt{490}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{490}}{2*4.9}=\frac{0+2\sqrt{490}}{9.8}
=\frac{2\sqrt{490}}{9.8}
$