H(t)=-16t^2+50t+110

Solution for H(t)=-16t^2+50t+110 equation:

(H)=-16H^2+50H+110

We move all terms to the left:

(H)-(-16H^2+50H+110)=0

We get rid of parentheses

16H^2-50H+H-110=0

We add all the numbers together, and all the variables

16H^2-49H-110=0

a = 16; b = -49; c = -110;
Δ = b2-4ac
Δ = -492-4·16·(-110)
Δ = 9441
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:

$\sqrt{\Delta}=\sqrt{9441}=\sqrt{9*1049}=\sqrt{9}*\sqrt{1049}=3\sqrt{1049}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-3\sqrt{1049}}{2*16}=\frac{49-3\sqrt{1049}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+3\sqrt{1049}}{2*16}=\frac{49+3\sqrt{1049}}{32}
$