F(t)=-16t^2+112t+288

Solution for F(t)=-16t^2+112t+288 equation:

(F)=-16F^2+112F+288

We move all terms to the left:

(F)-(-16F^2+112F+288)=0

We get rid of parentheses

16F^2-112F+F-288=0

We add all the numbers together, and all the variables

16F^2-111F-288=0

a = 16; b = -111; c = -288;
Δ = b2-4ac
Δ = -1112-4·16·(-288)
Δ = 30753
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{30753}=\sqrt{9*3417}=\sqrt{9}*\sqrt{3417}=3\sqrt{3417}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-111)-3\sqrt{3417}}{2*16}=\frac{111-3\sqrt{3417}}{32}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-111)+3\sqrt{3417}}{2*16}=\frac{111+3\sqrt{3417}}{32}
$