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 = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{5}}{2*-0.5}=\frac{-4-2\sqrt{5}}{-1}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{5}}{2*-0.5}=\frac{-4+2\sqrt{5}}{-1}
$