Solve 2x^2+2x-2

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

Factor

$$2\left(x-\frac{-\sqrt{5}-1}{2}\right)\left(x-\frac{\sqrt{5}-1}{2}\right)$$

Steps Using the Quadratic Formula

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.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

Square 2.

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

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-2±\sqrt{4+16}}{2\times 2}$$

Add 4 to 16.

$$x=\frac{-2±\sqrt{20}}{2\times 2}$$

Take the square root of 20.

$$x=\frac{-2±2\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±2\sqrt{5}}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{4} when ± is plus. Add -2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{4}$$

Divide -2+2\sqrt{5} by 4.

$$x=\frac{\sqrt{5}-1}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{4} when ± is minus. Subtract 2\sqrt{5} from -2.

$$x=\frac{-2\sqrt{5}-2}{4}$$

Divide -2-2\sqrt{5} by 4.

$$x=\frac{-\sqrt{5}-1}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{5}}{2} for x_{1} and \frac{-1-\sqrt{5}}{2} for x_{2}.

$$2x^{2}+2x-2=2\left(x-\frac{\sqrt{5}-1}{2}\right)\left(x-\frac{-\sqrt{5}-1}{2}\right)$$

Evaluate

$$2\left(x^{2}+x-1\right)$$