Solve 2/(6x-1)^2=10

$$\frac { 2 } { ( 6 x - 1 ) ^ { 2 } } = 10$$

Solve for x

$$x=-\frac{\sqrt{5}}{30}+\frac{1}{6}\approx 0.092131067$$

$$x=\frac{\sqrt{5}}{30}+\frac{1}{6}\approx 0.241202266$$

Steps Using the Quadratic Formula

Variable x cannot be equal to \frac{1}{6} since division by zero is not defined. Multiply both sides of the equation by \left(6x-1\right)^{2}.

$$2=10\left(6x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-1\right)^{2}.

$$2=10\left(36x^{2}-12x+1\right)$$

Use the distributive property to multiply 10 by 36x^{2}-12x+1.

$$2=360x^{2}-120x+10$$

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

$$360x^{2}-120x+10=2$$

Subtract 2 from both sides.

$$360x^{2}-120x+10-2=0$$

Subtract 2 from 10 to get 8.

$$360x^{2}-120x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 360 for a, -120 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-120\right)±\sqrt{\left(-120\right)^{2}-4\times 360\times 8}}{2\times 360}$$

Square -120.

$$x=\frac{-\left(-120\right)±\sqrt{14400-4\times 360\times 8}}{2\times 360}$$

Multiply -4 times 360.

$$x=\frac{-\left(-120\right)±\sqrt{14400-1440\times 8}}{2\times 360}$$

Multiply -1440 times 8.

$$x=\frac{-\left(-120\right)±\sqrt{14400-11520}}{2\times 360}$$

Add 14400 to -11520.

$$x=\frac{-\left(-120\right)±\sqrt{2880}}{2\times 360}$$

Take the square root of 2880.

$$x=\frac{-\left(-120\right)±24\sqrt{5}}{2\times 360}$$

The opposite of -120 is 120.

$$x=\frac{120±24\sqrt{5}}{2\times 360}$$

Multiply 2 times 360.

$$x=\frac{120±24\sqrt{5}}{720}$$

Now solve the equation x=\frac{120±24\sqrt{5}}{720} when ± is plus. Add 120 to 24\sqrt{5}.

$$x=\frac{24\sqrt{5}+120}{720}$$

Divide 120+24\sqrt{5} by 720.

$$x=\frac{\sqrt{5}}{30}+\frac{1}{6}$$

Now solve the equation x=\frac{120±24\sqrt{5}}{720} when ± is minus. Subtract 24\sqrt{5} from 120.

$$x=\frac{120-24\sqrt{5}}{720}$$

Divide 120-24\sqrt{5} by 720.

$$x=-\frac{\sqrt{5}}{30}+\frac{1}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{5}}{30}+\frac{1}{6} x=-\frac{\sqrt{5}}{30}+\frac{1}{6}$$

Steps for Completing the Square

Variable x cannot be equal to \frac{1}{6} since division by zero is not defined. Multiply both sides of the equation by \left(6x-1\right)^{2}.

$$2=10\left(6x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-1\right)^{2}.

$$2=10\left(36x^{2}-12x+1\right)$$

Use the distributive property to multiply 10 by 36x^{2}-12x+1.

$$2=360x^{2}-120x+10$$

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

$$360x^{2}-120x+10=2$$

Subtract 10 from both sides.

$$360x^{2}-120x=2-10$$

Subtract 10 from 2 to get -8.

$$360x^{2}-120x=-8$$

Divide both sides by 360.

$$\frac{360x^{2}-120x}{360}=-\frac{8}{360}$$

Dividing by 360 undoes the multiplication by 360.

$$x^{2}+\left(-\frac{120}{360}\right)x=-\frac{8}{360}$$

Reduce the fraction \frac{-120}{360} to lowest terms by extracting and canceling out 120.

$$x^{2}-\frac{1}{3}x=-\frac{8}{360}$$

Reduce the fraction \frac{-8}{360} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{1}{3}x=-\frac{1}{45}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=-\frac{1}{45}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{1}{45}+\frac{1}{36}$$

Add -\frac{1}{45} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{180}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{180}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{180}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{5}}{30} x-\frac{1}{6}=-\frac{\sqrt{5}}{30}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{\sqrt{5}}{30}+\frac{1}{6} x=-\frac{\sqrt{5}}{30}+\frac{1}{6}$$