H(t)=-16t^2+180t+10

Solution for H(t)=-16t^2+180t+10 equation:

(H)=-16H^2+180H+10

We move all terms to the left:

(H)-(-16H^2+180H+10)=0

We get rid of parentheses

16H^2-180H+H-10=0

We add all the numbers together, and all the variables

16H^2-179H-10=0

a = 16; b = -179; c = -10;
Δ = b2-4ac
Δ = -1792-4·16·(-10)
Δ = 32681
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{-(-179)-\sqrt{32681}}{2*16}=\frac{179-\sqrt{32681}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-179)+\sqrt{32681}}{2*16}=\frac{179+\sqrt{32681}}{32}
$