H(t)=350-16(t)^2

Solution for H(t)=350-16(t)^2 equation:

(H)=350-16(H)^2

We move all terms to the left:

(H)-(350-16(H)^2)=0

determiningTheFunctionDomain
-(350-16H^2)+H=0

We get rid of parentheses

16H^2+H-350=0

a = 16; b = 1; c = -350;
Δ = b2-4ac
Δ = 12-4·16·(-350)
Δ = 22401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{22401}=\sqrt{9*2489}=\sqrt{9}*\sqrt{2489}=3\sqrt{2489}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{2489}}{2*16}=\frac{-1-3\sqrt{2489}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{2489}}{2*16}=\frac{-1+3\sqrt{2489}}{32}
$