H=-7t(t-5)

Solution for H=-7t(t-5) equation:

=-7H(H-5)

We move all terms to the left:

-(-7H(H-5))=0


We calculate terms in parentheses: -(-7H(H-5)), so:

-7H(H-5)

We multiply parentheses

-7H^2+35H

Back to the equation:

-(-7H^2+35H)


We get rid of parentheses

7H^2-35H=0

a = 7; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·7·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*7}=\frac{0}{14}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*7}=\frac{70}{14}
=5
$