F(t)=-16t2+20t

Solution for F(t)=-16t2+20t equation:

(F)=-16F^2+20F

We move all terms to the left:

(F)-(-16F^2+20F)=0

We get rid of parentheses

16F^2-20F+F=0

We add all the numbers together, and all the variables

16F^2-19F=0

a = 16; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·16·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*16}=\frac{0}{32}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*16}=\frac{38}{32}
=1+3/16
$