H(t)=t^2+10t+12

Solution for H(t)=t^2+10t+12 equation:

(H)=H^2+10H+12

We move all terms to the left:

(H)-(H^2+10H+12)=0

We get rid of parentheses

-H^2+H-10H-12=0

We add all the numbers together, and all the variables

-1H^2-9H-12=0

a = -1; b = -9; c = -12;
Δ = b2-4ac
Δ = -92-4·(-1)·(-12)
Δ = 33
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{-(-9)-\sqrt{33}}{2*-1}=\frac{9-\sqrt{33}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{33}}{2*-1}=\frac{9+\sqrt{33}}{-2}
$