H=-16t^2+t+4

Solution for H=-16t^2+t+4 equation:

=-16H^2+H+4

We move all terms to the left:

-(-16H^2+H+4)=0

We get rid of parentheses

16H^2-H-4=0

We add all the numbers together, and all the variables

16H^2-1H-4=0

a = 16; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·16·(-4)
Δ = 257
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{-(-1)-\sqrt{257}}{2*16}=\frac{1-\sqrt{257}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{257}}{2*16}=\frac{1+\sqrt{257}}{32}
$