Solve 9x^2=21x

$$9 x ^ { 2 } = 21 x$$

Solve for x

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$x=0$$

Steps Using Factoring

Subtract 21x from both sides.

$$9x^{2}-21x=0$$

Factor out x.

$$x\left(9x-21\right)=0$$

To find equation solutions, solve x=0 and 9x-21=0.

$$x=0 x=\frac{7}{3}$$

Steps Using the Quadratic Formula

Subtract 21x from both sides.

$$9x^{2}-21x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -21 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}}}{2\times 9}$$

Take the square root of \left(-21\right)^{2}.

$$x=\frac{-\left(-21\right)±21}{2\times 9}$$

The opposite of -21 is 21.

$$x=\frac{21±21}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{21±21}{18}$$

Now solve the equation x=\frac{21±21}{18} when ± is plus. Add 21 to 21.

$$x=\frac{42}{18}$$

Reduce the fraction \frac{42}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{7}{3}$$

Now solve the equation x=\frac{21±21}{18} when ± is minus. Subtract 21 from 21.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

The equation is now solved.

$$x=\frac{7}{3} x=0$$

Steps for Completing the Square

Subtract 21x from both sides.

$$9x^{2}-21x=0$$

Divide both sides by 9.

$$\frac{9x^{2}-21x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{21}{9}\right)x=\frac{0}{9}$$

Reduce the fraction \frac{-21}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{7}{3}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}-\frac{7}{3}x=0$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{49}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{7}{6} x-\frac{7}{6}=-\frac{7}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=\frac{7}{3} x=0$$