H(t)=-16t2+300

Solution for H(t)=-16t2+300 equation:

(H)=-16H^2+300

We move all terms to the left:

(H)-(-16H^2+300)=0

We get rid of parentheses

16H^2+H-300=0

a = 16; b = 1; c = -300;
Δ = b2-4ac
Δ = 12-4·16·(-300)
Δ = 19201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{19201}}{2*16}=\frac{-1-\sqrt{19201}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{19201}}{2*16}=\frac{-1+\sqrt{19201}}{32}
$