H(2)=-16t2+95t

Solution for H(2)=-16t2+95t equation:

(2)=-16H^2+95H

We move all terms to the left:

(2)-(-16H^2+95H)=0

We get rid of parentheses

16H^2-95H+2=0

a = 16; b = -95; c = +2;
Δ = b2-4ac
Δ = -952-4·16·2
Δ = 8897
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{-(-95)-\sqrt{8897}}{2*16}=\frac{95-\sqrt{8897}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{8897}}{2*16}=\frac{95+\sqrt{8897}}{32}
$