H=-16t^2+140t+30

Solution for H=-16t^2+140t+30 equation:

=-16H^2+140H+30

We move all terms to the left:

-(-16H^2+140H+30)=0

We get rid of parentheses

16H^2-140H-30=0

a = 16; b = -140; c = -30;
Δ = b2-4ac
Δ = -1402-4·16·(-30)
Δ = 21520
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:

$\sqrt{\Delta}=\sqrt{21520}=\sqrt{16*1345}=\sqrt{16}*\sqrt{1345}=4\sqrt{1345}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-4\sqrt{1345}}{2*16}=\frac{140-4\sqrt{1345}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+4\sqrt{1345}}{2*16}=\frac{140+4\sqrt{1345}}{32}
$