H(t)=-16t^2+145t+80

Solution for H(t)=-16t^2+145t+80 equation:

(H)=-16H^2+145H+80

We move all terms to the left:

(H)-(-16H^2+145H+80)=0

We get rid of parentheses

16H^2-145H+H-80=0

We add all the numbers together, and all the variables

16H^2-144H-80=0

a = 16; b = -144; c = -80;
Δ = b2-4ac
Δ = -1442-4·16·(-80)
Δ = 25856
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:

$\sqrt{\Delta}=\sqrt{25856}=\sqrt{256*101}=\sqrt{256}*\sqrt{101}=16\sqrt{101}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-144)-16\sqrt{101}}{2*16}=\frac{144-16\sqrt{101}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-144)+16\sqrt{101}}{2*16}=\frac{144+16\sqrt{101}}{32}
$