H(t)=-3t^2+12t+60

Solution for H(t)=-3t^2+12t+60 equation:

(H)=-3H^2+12H+60

We move all terms to the left:

(H)-(-3H^2+12H+60)=0

We get rid of parentheses

3H^2-12H+H-60=0

We add all the numbers together, and all the variables

3H^2-11H-60=0

a = 3; b = -11; c = -60;
Δ = b2-4ac
Δ = -112-4·3·(-60)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-29}{2*3}=\frac{-18}{6}
=-3
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+29}{2*3}=\frac{40}{6}
=6+2/3
$