F(t)=0,5+3t-5t^2

Solution for F(t)=0,5+3t-5t^2 equation:

(F)=0.5+3F-5F^2

We move all terms to the left:

(F)-(0.5+3F-5F^2)=0

We get rid of parentheses

5F^2-3F+F-0.5=0

We add all the numbers together, and all the variables

5F^2-2F-0.5=0

a = 5; b = -2; c = -0.5;
Δ = b2-4ac
Δ = -22-4·5·(-0.5)
Δ = 14
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)-\sqrt{14}}{2*5}=\frac{2-\sqrt{14}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{14}}{2*5}=\frac{2+\sqrt{14}}{10}
$