H(t)=-4t^2+30t

Solution for H(t)=-4t^2+30t equation:

(H)=-4H^2+30H

We move all terms to the left:

(H)-(-4H^2+30H)=0

We get rid of parentheses

4H^2-30H+H=0

We add all the numbers together, and all the variables

4H^2-29H=0

a = 4; b = -29; c = 0;
Δ = b2-4ac
Δ = -292-4·4·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-29}{2*4}=\frac{0}{8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+29}{2*4}=\frac{58}{8}
=7+1/4
$