H(t)=-16t^2+48t+3.5

Solution for H(t)=-16t^2+48t+3.5 equation:

(H)=-16H^2+48H+3.5

We move all terms to the left:

(H)-(-16H^2+48H+3.5)=0

We get rid of parentheses

16H^2-48H+H-3.5=0

We add all the numbers together, and all the variables

16H^2-47H-3.5=0

a = 16; b = -47; c = -3.5;
Δ = b2-4ac
Δ = -472-4·16·(-3.5)
Δ = 2433
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{2433}}{2*16}=\frac{47-\sqrt{2433}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{2433}}{2*16}=\frac{47+\sqrt{2433}}{32}
$