H(t)=-12t^2+24t+3

Solution for H(t)=-12t^2+24t+3 equation:

(H)=-12H^2+24H+3

We move all terms to the left:

(H)-(-12H^2+24H+3)=0

We get rid of parentheses

12H^2-24H+H-3=0

We add all the numbers together, and all the variables

12H^2-23H-3=0

a = 12; b = -23; c = -3;
Δ = b2-4ac
Δ = -232-4·12·(-3)
Δ = 673
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{-(-23)-\sqrt{673}}{2*12}=\frac{23-\sqrt{673}}{24}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{673}}{2*12}=\frac{23+\sqrt{673}}{24}
$