F(x)=3x-0,2x^2+1.2

Solution for F(x)=3x-0,2x^2+1.2 equation:

(F)=3F-0.2F^2+1.2

We move all terms to the left:

(F)-(3F-0.2F^2+1.2)=0

We get rid of parentheses

0.2F^2-3F+F-1.2=0

We add all the numbers together, and all the variables

0.2F^2-2F-1.2=0

a = 0.2; b = -2; c = -1.2;
Δ = b2-4ac
Δ = -22-4·0.2·(-1.2)
Δ = 4.96
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{4.96}}{2*0.2}=\frac{2-\sqrt{4.96}}{0.4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{4.96}}{2*0.2}=\frac{2+\sqrt{4.96}}{0.4}
$