H(t)=-25t^2-5t+12

Solution for H(t)=-25t^2-5t+12 equation:

(H)=-25H^2-5H+12

We move all terms to the left:

(H)-(-25H^2-5H+12)=0

We get rid of parentheses

25H^2+5H+H-12=0

We add all the numbers together, and all the variables

25H^2+6H-12=0

a = 25; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·25·(-12)
Δ = 1236
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:

$\sqrt{\Delta}=\sqrt{1236}=\sqrt{4*309}=\sqrt{4}*\sqrt{309}=2\sqrt{309}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{309}}{2*25}=\frac{-6-2\sqrt{309}}{50}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{309}}{2*25}=\frac{-6+2\sqrt{309}}{50}
$