Solve 0=2x^2+6x+2

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

Solve for x

$$x=\frac{\sqrt{5}-3}{2}\approx -0.381966011$$

$$x=\frac{-\sqrt{5}-3}{2}\approx -2.618033989$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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.

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

The equation is now solved.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=-\frac{2}{2}$$

Divide 6 by 2.

$$x^{2}+3x=-\frac{2}{2}$$

Divide -2 by 2.

$$x^{2}+3x=-1$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-1+\frac{9}{4}$$

Add -1 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{5}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

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