Solve x^2-2x-1/6=x-x^2/3

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

Solve for x

$$x=\frac{\sqrt{2}}{4}+\frac{1}{2}\approx 0.853553391$$

$$x=-\frac{\sqrt{2}}{4}+\frac{1}{2}\approx 0.146446609$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 6,3.

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

To find the opposite of 2x-1, find the opposite of each term.

$$6x^{2}-2x+1=6x-2x^{2}$$

Subtract 6x from both sides.

$$6x^{2}-2x+1-6x=-2x^{2}$$

Combine -2x and -6x to get -8x.

$$6x^{2}-8x+1=-2x^{2}$$

Add 2x^{2} to both sides.

$$6x^{2}-8x+1+2x^{2}=0$$

Combine 6x^{2} and 2x^{2} to get 8x^{2}.

$$8x^{2}-8x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -8 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 8}}{2\times 8}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-8\right)±\sqrt{64-32}}{2\times 8}$$

Add 64 to -32.

$$x=\frac{-\left(-8\right)±\sqrt{32}}{2\times 8}$$

Take the square root of 32.

$$x=\frac{-\left(-8\right)±4\sqrt{2}}{2\times 8}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{2}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{8±4\sqrt{2}}{16}$$

Now solve the equation x=\frac{8±4\sqrt{2}}{16} when ± is plus. Add 8 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+8}{16}$$

Divide 8+4\sqrt{2} by 16.

$$x=\frac{\sqrt{2}}{4}+\frac{1}{2}$$

Now solve the equation x=\frac{8±4\sqrt{2}}{16} when ± is minus. Subtract 4\sqrt{2} from 8.

$$x=\frac{8-4\sqrt{2}}{16}$$

Divide 8-4\sqrt{2} by 16.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 6,3.

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

To find the opposite of 2x-1, find the opposite of each term.

$$6x^{2}-2x+1=6x-2x^{2}$$

Subtract 6x from both sides.

$$6x^{2}-2x+1-6x=-2x^{2}$$

Combine -2x and -6x to get -8x.

$$6x^{2}-8x+1=-2x^{2}$$

Add 2x^{2} to both sides.

$$6x^{2}-8x+1+2x^{2}=0$$

Combine 6x^{2} and 2x^{2} to get 8x^{2}.

$$8x^{2}-8x+1=0$$

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

$$8x^{2}-8x=-1$$

Divide both sides by 8.

$$\frac{8x^{2}-8x}{8}=-\frac{1}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{8}{8}\right)x=-\frac{1}{8}$$

Divide -8 by 8.

$$x^{2}-x=-\frac{1}{8}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{8}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=-\frac{1}{8}+\frac{1}{4}$$

Add -\frac{1}{8} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{1}{8}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{8}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{2}}{4} x-\frac{1}{2}=-\frac{\sqrt{2}}{4}$$

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

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