H(t)=-10t^2+60t+5

Solution for H(t)=-10t^2+60t+5 equation:

(H)=-10H^2+60H+5

We move all terms to the left:

(H)-(-10H^2+60H+5)=0

We get rid of parentheses

10H^2-60H+H-5=0

We add all the numbers together, and all the variables

10H^2-59H-5=0

a = 10; b = -59; c = -5;
Δ = b2-4ac
Δ = -592-4·10·(-5)
Δ = 3681
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:

$\sqrt{\Delta}=\sqrt{3681}=\sqrt{9*409}=\sqrt{9}*\sqrt{409}=3\sqrt{409}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-59)-3\sqrt{409}}{2*10}=\frac{59-3\sqrt{409}}{20}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-59)+3\sqrt{409}}{2*10}=\frac{59+3\sqrt{409}}{20}
$