H=60t-16t^2+15

Solution for H=60t-16t^2+15 equation:

=60H-16H^2+15

We move all terms to the left:

-(60H-16H^2+15)=0

We get rid of parentheses

16H^2-60H-15=0

a = 16; b = -60; c = -15;
Δ = b2-4ac
Δ = -602-4·16·(-15)
Δ = 4560
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:

$\sqrt{\Delta}=\sqrt{4560}=\sqrt{16*285}=\sqrt{16}*\sqrt{285}=4\sqrt{285}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{285}}{2*16}=\frac{60-4\sqrt{285}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{285}}{2*16}=\frac{60+4\sqrt{285}}{32}
$