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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1H^2+3H+15=0

a = -1; b = 3; c = +15;
Δ = b2-4ac
Δ = 32-4·(-1)·15
Δ = 69
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{-(3)-\sqrt{69}}{2*-1}=\frac{-3-\sqrt{69}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{69}}{2*-1}=\frac{-3+\sqrt{69}}{-2}
$