H=-4.9t^2+38t

Solution for H=-4.9t^2+38t equation:

=-4.9H^2+38H

We move all terms to the left:

-(-4.9H^2+38H)=0

We get rid of parentheses

4.9H^2-38H=0

a = 4.9; b = -38; c = 0;
Δ = b2-4ac
Δ = -382-4·4.9·0
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-38}{2*4.9}=\frac{0}{9.8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+38}{2*4.9}=\frac{76}{9.8}
=7+5.2857142857143/7
$