H(t)=-16t^2+42t+97

Solution for H(t)=-16t^2+42t+97 equation:

(H)=-16H^2+42H+97

We move all terms to the left:

(H)-(-16H^2+42H+97)=0

We get rid of parentheses

16H^2-42H+H-97=0

We add all the numbers together, and all the variables

16H^2-41H-97=0

a = 16; b = -41; c = -97;
Δ = b2-4ac
Δ = -412-4·16·(-97)
Δ = 7889
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:

$\sqrt{\Delta}=\sqrt{7889}=\sqrt{49*161}=\sqrt{49}*\sqrt{161}=7\sqrt{161}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-7\sqrt{161}}{2*16}=\frac{41-7\sqrt{161}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+7\sqrt{161}}{2*16}=\frac{41+7\sqrt{161}}{32}
$