Solve 2x^2+6x+2=

$$2 { x }^{ 2 } +6x+2=$$

Factor

$$2\left(x-\frac{-\sqrt{5}-3}{2}\right)\left(x-\frac{\sqrt{5}-3}{2}\right)$$

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.

$$2x^{2}+6x+2=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\times 2}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-6±\sqrt{36-16}}{2\times 2}$$

Add 36 to -16.

$$x=\frac{-6±\sqrt{20}}{2\times 2}$$

Take the square root of 20.

$$x=\frac{-6±2\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±2\sqrt{5}}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{4} when ± is plus. Add -6 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-6}{4}$$

Divide -6+2\sqrt{5} by 4.

$$x=\frac{\sqrt{5}-3}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{4} when ± is minus. Subtract 2\sqrt{5} from -6.

$$x=\frac{-2\sqrt{5}-6}{4}$$

Divide -6-2\sqrt{5} by 4.

$$x=\frac{-\sqrt{5}-3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{5}}{2} for x_{1} and \frac{-3-\sqrt{5}}{2} for x_{2}.

$$2x^{2}+6x+2=2\left(x-\frac{\sqrt{5}-3}{2}\right)\left(x-\frac{-\sqrt{5}-3}{2}\right)$$

Evaluate

$$2\left(x^{2}+3x+1\right)$$