F(3)=-16t^2+128t+144

Solution for F(3)=-16t^2+128t+144 equation:

(3)=-16F^2+128F+144

We move all terms to the left:

(3)-(-16F^2+128F+144)=0

We get rid of parentheses

16F^2-128F-144+3=0

We add all the numbers together, and all the variables

16F^2-128F-141=0

a = 16; b = -128; c = -141;
Δ = b2-4ac
Δ = -1282-4·16·(-141)
Δ = 25408
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:

$\sqrt{\Delta}=\sqrt{25408}=\sqrt{64*397}=\sqrt{64}*\sqrt{397}=8\sqrt{397}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-8\sqrt{397}}{2*16}=\frac{128-8\sqrt{397}}{32}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+8\sqrt{397}}{2*16}=\frac{128+8\sqrt{397}}{32}
$