H(t)=-16(t-8)(t+3)

Solution for H(t)=-16(t-8)(t+3) equation:

(H)=-16(H-8)(H+3)

We move all terms to the left:

(H)-(-16(H-8)(H+3))=0

We multiply parentheses ..

-(-16(+H^2+3H-8H-24))+H=0


We calculate terms in parentheses: -(-16(+H^2+3H-8H-24)), so:

-16(+H^2+3H-8H-24)

We multiply parentheses

-16H^2-48H+128H+384

We add all the numbers together, and all the variables

-16H^2+80H+384

Back to the equation:

-(-16H^2+80H+384)


We get rid of parentheses

16H^2-80H+H-384=0

We add all the numbers together, and all the variables

16H^2-79H-384=0

a = 16; b = -79; c = -384;
Δ = b2-4ac
Δ = -792-4·16·(-384)
Δ = 30817
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{-(-79)-\sqrt{30817}}{2*16}=\frac{79-\sqrt{30817}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79)+\sqrt{30817}}{2*16}=\frac{79+\sqrt{30817}}{32}
$