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

Solution for H(t)=-5t^2+15t+2 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5H^2-14H-2=0

a = 5; b = -14; c = -2;
Δ = b2-4ac
Δ = -142-4·5·(-2)
Δ = 236
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:

$\sqrt{\Delta}=\sqrt{236}=\sqrt{4*59}=\sqrt{4}*\sqrt{59}=2\sqrt{59}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{59}}{2*5}=\frac{14-2\sqrt{59}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{59}}{2*5}=\frac{14+2\sqrt{59}}{10}
$