Solve 2-2xleqx^2

$$2 - 2 x \leq x ^ { 2 }$$

Solve for x

$$x\in (-\infty,-\sqrt{3}-1]\cup [\sqrt{3}-1,\infty)$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$2-2x-x^{2}\leq 0$$

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

$$-2+2x+x^{2}\geq 0$$

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.

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

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

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

Do the calculations.

$$x=\frac{-2±2\sqrt{3}}{2}$$

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

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{3}-1\right)\right)\left(x-\left(-\sqrt{3}-1\right)\right)\geq 0$$

For the product to be ≥0, x-\left(\sqrt{3}-1\right) and x-\left(-\sqrt{3}-1\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(\sqrt{3}-1\right) and x-\left(-\sqrt{3}-1\right) are both ≤0.

$$x-\left(\sqrt{3}-1\right)\leq 0 x-\left(-\sqrt{3}-1\right)\leq 0$$

The solution satisfying both inequalities is x\leq -\left(\sqrt{3}+1\right).

$$x\leq -\left(\sqrt{3}+1\right)$$

Consider the case when x-\left(\sqrt{3}-1\right) and x-\left(-\sqrt{3}-1\right) are both ≥0.

$$x-\left(-\sqrt{3}-1\right)\geq 0 x-\left(\sqrt{3}-1\right)\geq 0$$

The solution satisfying both inequalities is x\geq \sqrt{3}-1.

$$x\geq \sqrt{3}-1$$

The final solution is the union of the obtained solutions.

$$x\leq -\sqrt{3}-1\text{; }x\geq \sqrt{3}-1$$