H(t)=-11(t-4)(t+1)

Solution for H(t)=-11(t-4)(t+1) equation:

(H)=-11(H-4)(H+1)

We move all terms to the left:

(H)-(-11(H-4)(H+1))=0

We multiply parentheses ..

-(-11(+H^2+H-4H-4))+H=0


We calculate terms in parentheses: -(-11(+H^2+H-4H-4)), so:

-11(+H^2+H-4H-4)

We multiply parentheses

-11H^2-11H+44H+44

We add all the numbers together, and all the variables

-11H^2+33H+44

Back to the equation:

-(-11H^2+33H+44)


We get rid of parentheses

11H^2-33H+H-44=0

We add all the numbers together, and all the variables

11H^2-32H-44=0

a = 11; b = -32; c = -44;
Δ = b2-4ac
Δ = -322-4·11·(-44)
Δ = 2960
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:

$\sqrt{\Delta}=\sqrt{2960}=\sqrt{16*185}=\sqrt{16}*\sqrt{185}=4\sqrt{185}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{185}}{2*11}=\frac{32-4\sqrt{185}}{22}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{185}}{2*11}=\frac{32+4\sqrt{185}}{22}
$