H(t)=15+12t-3t^2

Solution for H(t)=15+12t-3t^2 equation:

(H)=15+12H-3H^2

We move all terms to the left:

(H)-(15+12H-3H^2)=0

We get rid of parentheses

3H^2-12H+H-15=0

We add all the numbers together, and all the variables

3H^2-11H-15=0

a = 3; b = -11; c = -15;
Δ = b2-4ac
Δ = -112-4·3·(-15)
Δ = 301
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{-(-11)-\sqrt{301}}{2*3}=\frac{11-\sqrt{301}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{301}}{2*3}=\frac{11+\sqrt{301}}{6}
$