H=-16t^2+131t+347

Solution for H=-16t^2+131t+347 equation:

=-16H^2+131H+347

We move all terms to the left:

-(-16H^2+131H+347)=0

We get rid of parentheses

16H^2-131H-347=0

a = 16; b = -131; c = -347;
Δ = b2-4ac
Δ = -1312-4·16·(-347)
Δ = 39369
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{-(-131)-\sqrt{39369}}{2*16}=\frac{131-\sqrt{39369}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-131)+\sqrt{39369}}{2*16}=\frac{131+\sqrt{39369}}{32}
$