H(t)=-16t2+48t+64

Solution for H(t)=-16t2+48t+64 equation:

(H)=-16H^2+48H+64

We move all terms to the left:

(H)-(-16H^2+48H+64)=0

We get rid of parentheses

16H^2-48H+H-64=0

We add all the numbers together, and all the variables

16H^2-47H-64=0

a = 16; b = -47; c = -64;
Δ = b2-4ac
Δ = -472-4·16·(-64)
Δ = 6305
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{-(-47)-\sqrt{6305}}{2*16}=\frac{47-\sqrt{6305}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{6305}}{2*16}=\frac{47+\sqrt{6305}}{32}
$