Solve (15quad3x^2=27x

$$( 15 \quad 3 x ^ { 2 } = 27 x$$

Solve for x

$$x=\frac{3}{17}\approx 0.176470588$$

$$x=0$$

Steps Using Factoring

Subtract 27x from both sides.

$$153x^{2}-27x=0$$

Factor out x.

$$x\left(153x-27\right)=0$$

To find equation solutions, solve x=0 and 153x-27=0.

$$x=0 x=\frac{3}{17}$$

Steps Using the Quadratic Formula

Subtract 27x from both sides.

$$153x^{2}-27x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 153 for a, -27 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}}}{2\times 153}$$

Take the square root of \left(-27\right)^{2}.

$$x=\frac{-\left(-27\right)±27}{2\times 153}$$

The opposite of -27 is 27.

$$x=\frac{27±27}{2\times 153}$$

Multiply 2 times 153.

$$x=\frac{27±27}{306}$$

Now solve the equation x=\frac{27±27}{306} when ± is plus. Add 27 to 27.

$$x=\frac{54}{306}$$

Reduce the fraction \frac{54}{306} to lowest terms by extracting and canceling out 18.

$$x=\frac{3}{17}$$

Now solve the equation x=\frac{27±27}{306} when ± is minus. Subtract 27 from 27.

$$x=\frac{0}{306}$$

Divide 0 by 306.

$$x=0$$

The equation is now solved.

$$x=\frac{3}{17} x=0$$

Steps for Completing the Square

Subtract 27x from both sides.

$$153x^{2}-27x=0$$

Divide both sides by 153.

$$\frac{153x^{2}-27x}{153}=\frac{0}{153}$$

Dividing by 153 undoes the multiplication by 153.

$$x^{2}+\left(-\frac{27}{153}\right)x=\frac{0}{153}$$

Reduce the fraction \frac{-27}{153} to lowest terms by extracting and canceling out 9.

$$x^{2}-\frac{3}{17}x=\frac{0}{153}$$

Divide 0 by 153.

$$x^{2}-\frac{3}{17}x=0$$

Divide -\frac{3}{17}, the coefficient of the x term, by 2 to get -\frac{3}{34}. Then add the square of -\frac{3}{34} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{17}x+\left(-\frac{3}{34}\right)^{2}=\left(-\frac{3}{34}\right)^{2}$$

Square -\frac{3}{34} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{17}x+\frac{9}{1156}=\frac{9}{1156}$$

Factor x^{2}-\frac{3}{17}x+\frac{9}{1156}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{34}\right)^{2}=\frac{9}{1156}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{34}\right)^{2}}=\sqrt{\frac{9}{1156}}$$

Simplify.

$$x-\frac{3}{34}=\frac{3}{34} x-\frac{3}{34}=-\frac{3}{34}$$

Add \frac{3}{34} to both sides of the equation.

$$x=\frac{3}{17} x=0$$