Solve 2x(1-x)+1-x<0

$$2 x ( 1 - x ) + 1 - x < 0$$

Solve for x

$$x\in \left(-\infty,-\frac{1}{2}\right)\cup \left(1,\infty\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by 1-x.

$$2x-2x^{2}+1-x<0$$

Combine 2x and -x to get x.

$$x-2x^{2}+1<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in x-2x^{2}+1 positive. Since -1 is negative, the inequality direction is changed.

$$-x+2x^{2}-1>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-x+2x^{2}-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 2 for a, -1 for b, and -1 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{1±3}{4}$$

Solve the equation x=\frac{1±3}{4} when ± is plus and when ± is minus.

$$x=1 x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-1\right)\left(x+\frac{1}{2}\right)>0$$

For the product to be positive, x-1 and x+\frac{1}{2} have to be both negative or both positive. Consider the case when x-1 and x+\frac{1}{2} are both negative.

$$x-1<0 x+\frac{1}{2}<0$$

The solution satisfying both inequalities is x<-\frac{1}{2}.

$$x<-\frac{1}{2}$$

Consider the case when x-1 and x+\frac{1}{2} are both positive.

$$x+\frac{1}{2}>0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{2}\text{; }x>1$$