H(t)=-5t2+5t+5

Solution for H(t)=-5t2+5t+5 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5H^2-4H-5=0

a = 5; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·5·(-5)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{29}}{2*5}=\frac{4-2\sqrt{29}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{29}}{2*5}=\frac{4+2\sqrt{29}}{10}
$