H(t)=t2-5t

Solution for H(t)=t2-5t equation:

(H)=H2-5H

We move all terms to the left:

(H)-(H2-5H)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

-H^2+5H+H=0

We add all the numbers together, and all the variables

-1H^2+6H=0

a = -1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-1)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-1}=\frac{-12}{-2}
=+6
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-1}=\frac{0}{-2}
=0
$