F(t)=-4.9t^2+4.9t+5

Solution for F(t)=-4.9t^2+4.9t+5 equation:

(F)=-4.9F^2+4.9F+5

We move all terms to the left:

(F)-(-4.9F^2+4.9F+5)=0

We get rid of parentheses

4.9F^2-4.9F+F-5=0

We add all the numbers together, and all the variables

4.9F^2-3.9F-5=0

a = 4.9; b = -3.9; c = -5;
Δ = b2-4ac
Δ = -3.92-4·4.9·(-5)
Δ = 113.21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3.9)-\sqrt{113.21}}{2*4.9}=\frac{3.9-\sqrt{113.21}}{9.8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.9)+\sqrt{113.21}}{2*4.9}=\frac{3.9+\sqrt{113.21}}{9.8}
$