H(t)=-16t^2+140t+3

Solution for H(t)=-16t^2+140t+3 equation:

(H)=-16H^2+140H+3

We move all terms to the left:

(H)-(-16H^2+140H+3)=0

We get rid of parentheses

16H^2-140H+H-3=0

We add all the numbers together, and all the variables

16H^2-139H-3=0

a = 16; b = -139; c = -3;
Δ = b2-4ac
Δ = -1392-4·16·(-3)
Δ = 19513
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{-(-139)-\sqrt{19513}}{2*16}=\frac{139-\sqrt{19513}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-139)+\sqrt{19513}}{2*16}=\frac{139+\sqrt{19513}}{32}
$