H(t)=-16t^2+17t+3

Solution for H(t)=-16t^2+17t+3 equation:

(H)=-16H^2+17H+3

We move all terms to the left:

(H)-(-16H^2+17H+3)=0

We get rid of parentheses

16H^2-17H+H-3=0

We add all the numbers together, and all the variables

16H^2-16H-3=0

a = 16; b = -16; c = -3;
Δ = b2-4ac
Δ = -162-4·16·(-3)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{7}}{2*16}=\frac{16-8\sqrt{7}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{7}}{2*16}=\frac{16+8\sqrt{7}}{32}
$