F(x)=x(4x-2)+x

Solution for F(x)=x(4x-2)+x equation:

(F)=F(4F-2)+F

We move all terms to the left:

(F)-(F(4F-2)+F)=0


We calculate terms in parentheses: -(F(4F-2)+F), so:

F(4F-2)+F

We add all the numbers together, and all the variables

F+F(4F-2)

We multiply parentheses

4F^2+F-2F

We add all the numbers together, and all the variables

4F^2-1F

Back to the equation:

-(4F^2-1F)


We get rid of parentheses

-4F^2+F+1F=0

We add all the numbers together, and all the variables

-4F^2+2F=0

a = -4; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-4)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-4}=\frac{-4}{-8}
=1/2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-4}=\frac{0}{-8}
=0
$