H=-16t^2+62t+I

Solution for H=-16t^2+62t+I equation:

=-16H^2+62H+

We move all terms to the left:

-(-16H^2+62H+)=0

We get rid of parentheses

16H^2-62H-=0

We add all the numbers together, and all the variables

16H^2-62H=0

a = 16; b = -62; c = 0;
Δ = b2-4ac
Δ = -622-4·16·0
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-62}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+62}{2*16}=\frac{124}{32}
=3+7/8
$