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

Solution for F(x)=0.5x^2-3x+2 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-0.5F^2+4F-2=0

a = -0.5; b = 4; c = -2;
Δ = b2-4ac
Δ = 42-4·(-0.5)·(-2)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{3}}{2*-0.5}=\frac{-4-2\sqrt{3}}{-1}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{3}}{2*-0.5}=\frac{-4+2\sqrt{3}}{-1}
$