Solve (2)3x^2-6x-2=0

$$( 2 ) 3 x ^ { 2 } - 6 x - 2 = 0$$

Solve for x

$$x=\frac{\sqrt{21}}{6}+\frac{1}{2}\approx 1.263762616$$

$$x=-\frac{\sqrt{21}}{6}+\frac{1}{2}\approx -0.263762616$$

Steps Using the Quadratic Formula

Multiply 2 and 3 to get 6.

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

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

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 6\left(-2\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-6\right)±\sqrt{36-24\left(-2\right)}}{2\times 6}$$

Multiply -24 times -2.

$$x=\frac{-\left(-6\right)±\sqrt{36+48}}{2\times 6}$$

Add 36 to 48.

$$x=\frac{-\left(-6\right)±\sqrt{84}}{2\times 6}$$

Take the square root of 84.

$$x=\frac{-\left(-6\right)±2\sqrt{21}}{2\times 6}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{21}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{6±2\sqrt{21}}{12}$$

Now solve the equation x=\frac{6±2\sqrt{21}}{12} when ± is plus. Add 6 to 2\sqrt{21}.

$$x=\frac{2\sqrt{21}+6}{12}$$

Divide 6+2\sqrt{21} by 12.

$$x=\frac{\sqrt{21}}{6}+\frac{1}{2}$$

Now solve the equation x=\frac{6±2\sqrt{21}}{12} when ± is minus. Subtract 2\sqrt{21} from 6.

$$x=\frac{6-2\sqrt{21}}{12}$$

Divide 6-2\sqrt{21} by 12.

$$x=-\frac{\sqrt{21}}{6}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}}{6}+\frac{1}{2} x=-\frac{\sqrt{21}}{6}+\frac{1}{2}$$

Steps for Completing the Square

Multiply 2 and 3 to get 6.

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

Add 2 to both sides. Anything plus zero gives itself.

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

Divide both sides by 6.

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

Dividing by 6 undoes the multiplication by 6.

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

Divide -6 by 6.

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

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{1}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{7}{12}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{7}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{7}{12}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{21}}{6} x-\frac{1}{2}=-\frac{\sqrt{21}}{6}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{21}}{6}+\frac{1}{2} x=-\frac{\sqrt{21}}{6}+\frac{1}{2}$$