Solve (y^2+2y-2)

$$( y ^ { 2 } + 2 y - 2 )$$

Factor

$$\left(y-\left(-\sqrt{3}-1\right)\right)\left(y-\left(\sqrt{3}-1\right)\right)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times -2.

$$y=\frac{-2±\sqrt{4+8}}{2}$$

Add 4 to 8.

$$y=\frac{-2±\sqrt{12}}{2}$$

Take the square root of 12.

$$y=\frac{-2±2\sqrt{3}}{2}$$

Now solve the equation y=\frac{-2±2\sqrt{3}}{2} when ± is plus. Add -2 to 2\sqrt{3}.

$$y=\frac{2\sqrt{3}-2}{2}$$

Divide -2+2\sqrt{3} by 2.

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

Now solve the equation y=\frac{-2±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from -2.

$$y=\frac{-2\sqrt{3}-2}{2}$$

Divide -2-2\sqrt{3} by 2.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1+\sqrt{3} for x_{1} and -1-\sqrt{3} for x_{2}.

$$y^{2}+2y-2=\left(y-\left(\sqrt{3}-1\right)\right)\left(y-\left(-\sqrt{3}-1\right)\right)$$

Evaluate

$$y^{2}+2y-2$$