Solve left(7xright)^2-14x=0

$${ \left(7x \right) }^{ 2 } -14x = 0$$

Solve for x

$$x=\frac{2}{7}\approx 0.285714286$$

$$x=0$$

Steps Using Factoring

Expand \left(7x\right)^{2}.

$$7^{2}x^{2}-14x=0$$

Calculate 7 to the power of 2 and get 49.

$$49x^{2}-14x=0$$

Factor out x.

$$x\left(49x-14\right)=0$$

To find equation solutions, solve x=0 and 49x-14=0.

$$x=0 x=\frac{2}{7}$$

Steps Using the Quadratic Formula

Expand \left(7x\right)^{2}.

$$7^{2}x^{2}-14x=0$$

Calculate 7 to the power of 2 and get 49.

$$49x^{2}-14x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -14 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}}}{2\times 49}$$

Take the square root of \left(-14\right)^{2}.

$$x=\frac{-\left(-14\right)±14}{2\times 49}$$

The opposite of -14 is 14.

$$x=\frac{14±14}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{14±14}{98}$$

Now solve the equation x=\frac{14±14}{98} when ± is plus. Add 14 to 14.

$$x=\frac{28}{98}$$

Reduce the fraction \frac{28}{98} to lowest terms by extracting and canceling out 14.

$$x=\frac{2}{7}$$

Now solve the equation x=\frac{14±14}{98} when ± is minus. Subtract 14 from 14.

$$x=\frac{0}{98}$$

Divide 0 by 98.

$$x=0$$

The equation is now solved.

$$x=\frac{2}{7} x=0$$

Steps for Completing the Square

Expand \left(7x\right)^{2}.

$$7^{2}x^{2}-14x=0$$

Calculate 7 to the power of 2 and get 49.

$$49x^{2}-14x=0$$

Divide both sides by 49.

$$\frac{49x^{2}-14x}{49}=\frac{0}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$x^{2}+\left(-\frac{14}{49}\right)x=\frac{0}{49}$$

Reduce the fraction \frac{-14}{49} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{2}{7}x=\frac{0}{49}$$

Divide 0 by 49.

$$x^{2}-\frac{2}{7}x=0$$

Divide -\frac{2}{7}, the coefficient of the x term, by 2 to get -\frac{1}{7}. Then add the square of -\frac{1}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{7}x+\frac{1}{49}=\frac{1}{49}$$

Factor x^{2}-\frac{2}{7}x+\frac{1}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{7}\right)^{2}=\frac{1}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{7}\right)^{2}}=\sqrt{\frac{1}{49}}$$

Simplify.

$$x-\frac{1}{7}=\frac{1}{7} x-\frac{1}{7}=-\frac{1}{7}$$

Add \frac{1}{7} to both sides of the equation.

$$x=\frac{2}{7} x=0$$