F(x)=0.3x^2+1.2x

Solution for F(x)=0.3x^2+1.2x equation:

(F)=0.3F^2+1.2F

We move all terms to the left:

(F)-(0.3F^2+1.2F)=0

We get rid of parentheses

-0.3F^2+F-1.2F=0

We add all the numbers together, and all the variables

-0.3F^2-0.2F=0

a = -0.3; b = -0.2; c = 0;
Δ = b2-4ac
Δ = -0.22-4·(-0.3)·0
Δ = 0.04
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{-(-0.2)-\sqrt{0.04}}{2*-0.3}=\frac{0.2-\sqrt{0.04}}{-0.6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.2)+\sqrt{0.04}}{2*-0.3}=\frac{0.2+\sqrt{0.04}}{-0.6}
$