H(t)=t^2+19t+14

Solution for H(t)=t^2+19t+14 equation:

(H)=H^2+19H+14

We move all terms to the left:

(H)-(H^2+19H+14)=0

We get rid of parentheses

-H^2+H-19H-14=0

We add all the numbers together, and all the variables

-1H^2-18H-14=0

a = -1; b = -18; c = -14;
Δ = b2-4ac
Δ = -182-4·(-1)·(-14)
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{67}}{2*-1}=\frac{18-2\sqrt{67}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{67}}{2*-1}=\frac{18+2\sqrt{67}}{-2}
$