H(t)=-5t^2+10+15

Solution for H(t)=-5t^2+10+15 equation:

(H)=-5H^2+10+15

We move all terms to the left:

(H)-(-5H^2+10+15)=0

We get rid of parentheses

5H^2+H-10-15=0

We add all the numbers together, and all the variables

5H^2+H-25=0

a = 5; b = 1; c = -25;
Δ = b2-4ac
Δ = 12-4·5·(-25)
Δ = 501
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{501}}{2*5}=\frac{-1-\sqrt{501}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{501}}{2*5}=\frac{-1+\sqrt{501}}{10}
$