H(t)=35t-5t^2

Solution for H(t)=35t-5t^2 equation:

(H)=35H-5H^2

We move all terms to the left:

(H)-(35H-5H^2)=0

We get rid of parentheses

5H^2-35H+H=0

We add all the numbers together, and all the variables

5H^2-34H=0

a = 5; b = -34; c = 0;
Δ = b2-4ac
Δ = -342-4·5·0
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-34}{2*5}=\frac{0}{10}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+34}{2*5}=\frac{68}{10}
=6+4/5
$