H(t)=-5t^2+5t+30

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

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

We move all terms to the left:

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

We get rid of parentheses

5H^2-5H+H-30=0

We add all the numbers together, and all the variables

5H^2-4H-30=0

a = 5; b = -4; c = -30;
Δ = b2-4ac
Δ = -42-4·5·(-30)
Δ = 616
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:

$\sqrt{\Delta}=\sqrt{616}=\sqrt{4*154}=\sqrt{4}*\sqrt{154}=2\sqrt{154}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{154}}{2*5}=\frac{4-2\sqrt{154}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{154}}{2*5}=\frac{4+2\sqrt{154}}{10}
$