-16t^2+150t+110=H(t)

Solution for -16t^2+150t+110=H(t) equation:

-16t^2+150t+110=(t)

We move all terms to the left:

-16t^2+150t+110-((t))=0

determiningTheFunctionDomain
-16t^2+150t-t+110=0

We add all the numbers together, and all the variables

-16t^2+149t+110=0

a = -16; b = 149; c = +110;
Δ = b2-4ac
Δ = 1492-4·(-16)·110
Δ = 29241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{29241}=171$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(149)-171}{2*-16}=\frac{-320}{-32}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(149)+171}{2*-16}=\frac{22}{-32}
=-11/16
$