F(t)=-16t^2+48t

Solution for F(t)=-16t^2+48t equation:

(F)=-16F^2+48F

We move all terms to the left:

(F)-(-16F^2+48F)=0

We get rid of parentheses

16F^2-48F+F=0

We add all the numbers together, and all the variables

16F^2-47F=0

a = 16; b = -47; c = 0;
Δ = b2-4ac
Δ = -472-4·16·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-47}{2*16}=\frac{0}{32}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+47}{2*16}=\frac{94}{32}
=2+15/16
$