H(t)=-4t^2+32t+1

Solution for H(t)=-4t^2+32t+1 equation:

(H)=-4H^2+32H+1

We move all terms to the left:

(H)-(-4H^2+32H+1)=0

We get rid of parentheses

4H^2-32H+H-1=0

We add all the numbers together, and all the variables

4H^2-31H-1=0

a = 4; b = -31; c = -1;
Δ = b2-4ac
Δ = -312-4·4·(-1)
Δ = 977
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{-(-31)-\sqrt{977}}{2*4}=\frac{31-\sqrt{977}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{977}}{2*4}=\frac{31+\sqrt{977}}{8}
$