H(t)=16t^2+150t

Solution for H(t)=16t^2+150t equation:

(H)=16H^2+150H

We move all terms to the left:

(H)-(16H^2+150H)=0

We get rid of parentheses

-16H^2+H-150H=0

We add all the numbers together, and all the variables

-16H^2-149H=0

a = -16; b = -149; c = 0;
Δ = b2-4ac
Δ = -1492-4·(-16)·0
Δ = 22201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{22201}=149$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-149)-149}{2*-16}=\frac{0}{-32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-149)+149}{2*-16}=\frac{298}{-32}
=-9+5/16
$