H(t)=-5t2+15t+12

Solution for H(t)=-5t2+15t+12 equation:

(H)=-5H^2+15H+12

We move all terms to the left:

(H)-(-5H^2+15H+12)=0

We get rid of parentheses

5H^2-15H+H-12=0

We add all the numbers together, and all the variables

5H^2-14H-12=0

a = 5; b = -14; c = -12;
Δ = b2-4ac
Δ = -142-4·5·(-12)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{109}}{2*5}=\frac{14-2\sqrt{109}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{109}}{2*5}=\frac{14+2\sqrt{109}}{10}
$