Solve 0=2left(1-3xright)^2-1

$$0=2 { \left(1-3x \right) }^{ 2 } -1$$

Solve for x

$$x=-\frac{\sqrt{2}}{6}+\frac{1}{3}\approx 0.097631073$$

$$x=\frac{\sqrt{2}}{6}+\frac{1}{3}\approx 0.569035594$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-3x\right)^{2}.

$$0=2\left(1-6x+9x^{2}\right)-1$$

Use the distributive property to multiply 2 by 1-6x+9x^{2}.

$$0=2-12x+18x^{2}-1$$

Subtract 1 from 2 to get 1.

$$0=1-12x+18x^{2}$$

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

$$1-12x+18x^{2}=0$$

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.

$$18x^{2}-12x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -12 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 18}}{2\times 18}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 18}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-12\right)±\sqrt{144-72}}{2\times 18}$$

Add 144 to -72.

$$x=\frac{-\left(-12\right)±\sqrt{72}}{2\times 18}$$

Take the square root of 72.

$$x=\frac{-\left(-12\right)±6\sqrt{2}}{2\times 18}$$

The opposite of -12 is 12.

$$x=\frac{12±6\sqrt{2}}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{12±6\sqrt{2}}{36}$$

Now solve the equation x=\frac{12±6\sqrt{2}}{36} when ± is plus. Add 12 to 6\sqrt{2}.

$$x=\frac{6\sqrt{2}+12}{36}$$

Divide 12+6\sqrt{2} by 36.

$$x=\frac{\sqrt{2}}{6}+\frac{1}{3}$$

Now solve the equation x=\frac{12±6\sqrt{2}}{36} when ± is minus. Subtract 6\sqrt{2} from 12.

$$x=\frac{12-6\sqrt{2}}{36}$$

Divide 12-6\sqrt{2} by 36.

$$x=-\frac{\sqrt{2}}{6}+\frac{1}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{2}}{6}+\frac{1}{3} x=-\frac{\sqrt{2}}{6}+\frac{1}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-3x\right)^{2}.

$$0=2\left(1-6x+9x^{2}\right)-1$$

Use the distributive property to multiply 2 by 1-6x+9x^{2}.

$$0=2-12x+18x^{2}-1$$

Subtract 1 from 2 to get 1.

$$0=1-12x+18x^{2}$$

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

$$1-12x+18x^{2}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-12x+18x^{2}=-1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$18x^{2}-12x=-1$$

Divide both sides by 18.

$$\frac{18x^{2}-12x}{18}=-\frac{1}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\left(-\frac{12}{18}\right)x=-\frac{1}{18}$$

Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{2}{3}x=-\frac{1}{18}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{1}{18}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{1}{18}+\frac{1}{9}$$

Add -\frac{1}{18} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{1}{18}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{1}{18}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{1}{18}}$$

Simplify.

$$x-\frac{1}{3}=\frac{\sqrt{2}}{6} x-\frac{1}{3}=-\frac{\sqrt{2}}{6}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{\sqrt{2}}{6}+\frac{1}{3} x=-\frac{\sqrt{2}}{6}+\frac{1}{3}$$