H=-16t^2+22t+140

Solution for H=-16t^2+22t+140 equation:

=-16H^2+22H+140

We move all terms to the left:

-(-16H^2+22H+140)=0

We get rid of parentheses

16H^2-22H-140=0

a = 16; b = -22; c = -140;
Δ = b2-4ac
Δ = -222-4·16·(-140)
Δ = 9444
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:

$\sqrt{\Delta}=\sqrt{9444}=\sqrt{4*2361}=\sqrt{4}*\sqrt{2361}=2\sqrt{2361}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{2361}}{2*16}=\frac{22-2\sqrt{2361}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{2361}}{2*16}=\frac{22+2\sqrt{2361}}{32}
$