H(t)=t^2+6t

Solution for H(t)=t^2+6t equation:

(H)=H^2+6H

We move all terms to the left:

(H)-(H^2+6H)=0

We get rid of parentheses

-H^2+H-6H=0

We add all the numbers together, and all the variables

-1H^2-5H=0

a = -1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-1)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-1}=\frac{0}{-2}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-1}=\frac{10}{-2}
=-5
$