H=-5t^2+30t+13

Solution for H=-5t^2+30t+13 equation:

=-5H^2+30H+13

We move all terms to the left:

-(-5H^2+30H+13)=0

We get rid of parentheses

5H^2-30H-13=0

a = 5; b = -30; c = -13;
Δ = b2-4ac
Δ = -302-4·5·(-13)
Δ = 1160
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:

$\sqrt{\Delta}=\sqrt{1160}=\sqrt{4*290}=\sqrt{4}*\sqrt{290}=2\sqrt{290}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{290}}{2*5}=\frac{30-2\sqrt{290}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{290}}{2*5}=\frac{30+2\sqrt{290}}{10}
$