Solve 2x-2<x^2-x

$$2x-2 < { x }^{ 2 } -x$$

Solve for x

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

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$2x-2-x^{2}<-x$$

Add x to both sides.

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

Combine 2x and x to get 3x.

$$3x-2-x^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 3x-2-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-3x+2+x^{2}>0$$

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.

$$-3x+2+x^{2}=0$$

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 1 for a, -3 for b, and 2 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 2}}{2}$$

Do the calculations.

$$x=\frac{3±1}{2}$$

Solve the equation x=\frac{3±1}{2} when ± is plus and when ± is minus.

$$x=2 x=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-2\right)\left(x-1\right)>0$$

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

$$x-2<0 x-1<0$$

The solution satisfying both inequalities is x<1.

$$x<1$$

Consider the case when x-2 and x-1 are both positive.

$$x-1>0 x-2>0$$

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

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