H=-16t2+150t+3

Solution for H=-16t2+150t+3 equation:

=-16H^2+150H+3

We move all terms to the left:

-(-16H^2+150H+3)=0

We get rid of parentheses

16H^2-150H-3=0

a = 16; b = -150; c = -3;
Δ = b2-4ac
Δ = -1502-4·16·(-3)
Δ = 22692
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:

$\sqrt{\Delta}=\sqrt{22692}=\sqrt{4*5673}=\sqrt{4}*\sqrt{5673}=2\sqrt{5673}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-150)-2\sqrt{5673}}{2*16}=\frac{150-2\sqrt{5673}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-150)+2\sqrt{5673}}{2*16}=\frac{150+2\sqrt{5673}}{32}
$