F(x)=1-3x+x^2/1+x+x^2

Solution for F(x)=1-3x+x^2/1+x+x^2 equation:

(F)=1-3F+F^2/1+F+F^2

We move all terms to the left:

(F)-(1-3F+F^2/1+F+F^2)=0


Domain of the equation: 1+F+F^2)!=0

We move all terms containing F to the left, all other terms to the right

F^2)+F!=-1

F∈R


We get rid of parentheses

-F^2/1-F^2+3F-F+F-1=0

We multiply all the terms by the denominator


-F^2-F^2*1+3F*1-F*1+F*1-1*1=0

We add all the numbers together, and all the variables

-1F^2-F^2*1+3F*1-F*1+F*1-1=0

Wy multiply elements

-1F^2-1F^2+3F-1F+F-1=0

We add all the numbers together, and all the variables

-2F^2+3F-1=0

a = -2; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·(-2)·(-1)
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{-(3)-1}{2*-2}=\frac{-4}{-4}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*-2}=\frac{-2}{-4}
=1/2
$