F(t)=0.2t^2+2t

Solution for F(t)=0.2t^2+2t equation:

(F)=0.2F^2+2F

We move all terms to the left:

(F)-(0.2F^2+2F)=0

We get rid of parentheses

-0.2F^2+F-2F=0

We add all the numbers together, and all the variables

-0.2F^2-1F=0

a = -0.2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-0.2)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-0.2}=\frac{0}{-0.4}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-0.2}=\frac{2}{-0.4}
=-5
$