F(t)=-16t^2+104t

Solution for F(t)=-16t^2+104t equation:

(F)=-16F^2+104F

We move all terms to the left:

(F)-(-16F^2+104F)=0

We get rid of parentheses

16F^2-104F+F=0

We add all the numbers together, and all the variables

16F^2-103F=0

a = 16; b = -103; c = 0;
Δ = b2-4ac
Δ = -1032-4·16·0
Δ = 10609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10609}=103$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-103)-103}{2*16}=\frac{0}{32}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-103)+103}{2*16}=\frac{206}{32}
=6+7/16
$