H(t)=-16t^2+60t+0

Solution for H(t)=-16t^2+60t+0 equation:

(H)=-16H^2+60H+0

We move all terms to the left:

(H)-(-16H^2+60H+0)=0

We get rid of parentheses

16H^2-60H+H-0=0

We add all the numbers together, and all the variables

16H^2-59H=0

a = 16; b = -59; c = 0;
Δ = b2-4ac
Δ = -592-4·16·0
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-59)-59}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-59)+59}{2*16}=\frac{118}{32}
=3+11/16
$