F(t)=15+3.5t-4.9t^2

Solution for F(t)=15+3.5t-4.9t^2 equation:

(F)=15+3.5F-4.9F^2

We move all terms to the left:

(F)-(15+3.5F-4.9F^2)=0

We get rid of parentheses

4.9F^2-3.5F+F-15=0

We add all the numbers together, and all the variables

4.9F^2-2.5F-15=0

a = 4.9; b = -2.5; c = -15;
Δ = b2-4ac
Δ = -2.52-4·4.9·(-15)
Δ = 300.25
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{-(-2.5)-\sqrt{300.25}}{2*4.9}=\frac{2.5-\sqrt{300.25}}{9.8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.5)+\sqrt{300.25}}{2*4.9}=\frac{2.5+\sqrt{300.25}}{9.8}
$