Solve 14x^2=7x

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

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=0$$

Steps Using Factoring

Subtract 7x from both sides.

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

Factor out x.

$$x\left(14x-7\right)=0$$

To find equation solutions, solve x=0 and 14x-7=0.

$$x=0 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 14 for a, -7 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}}}{2\times 14}$$

Take the square root of \left(-7\right)^{2}.

$$x=\frac{-\left(-7\right)±7}{2\times 14}$$

The opposite of -7 is 7.

$$x=\frac{7±7}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{7±7}{28}$$

Now solve the equation x=\frac{7±7}{28} when ± is plus. Add 7 to 7.

$$x=\frac{14}{28}$$

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

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{7±7}{28} when ± is minus. Subtract 7 from 7.

$$x=\frac{0}{28}$$

Divide 0 by 28.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{2} x=0$$

Steps for Completing the Square

Subtract 7x from both sides.

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

Divide both sides by 14.

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

Dividing by 14 undoes the multiplication by 14.

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

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

$$x^{2}-\frac{1}{2}x=\frac{0}{14}$$

Divide 0 by 14.

$$x^{2}-\frac{1}{2}x=0$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

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

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1}{2} x=0$$