Solve {l}{x/2=15x^2/3}{text{find}}{x}

$$\left. \begin{array} { l } { \frac { x } { 2 } = \frac { 15 x ^ { 2 } } { 3 } } \\ { \text { find } } \\ { x } \end{array} \right.$$

Solve for x

$$x=\frac{1}{10}=0.1$$

$$x=0$$

Steps Using Factoring

Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x=2\times 15x^{2}$$

Multiply 2 and 15 to get 30.

$$3x=30x^{2}$$

Subtract 30x^{2} from both sides.

$$3x-30x^{2}=0$$

Factor out x.

$$x\left(3-30x\right)=0$$

To find equation solutions, solve x=0 and 3-30x=0.

$$x=0 x=\frac{1}{10}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x=2\times 15x^{2}$$

Multiply 2 and 15 to get 30.

$$3x=30x^{2}$$

Subtract 30x^{2} from both sides.

$$3x-30x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-30x^{2}+3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -30 for a, 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}}}{2\left(-30\right)}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\left(-30\right)}$$

Multiply 2 times -30.

$$x=\frac{-3±3}{-60}$$

Now solve the equation x=\frac{-3±3}{-60} when ± is plus. Add -3 to 3.

$$x=\frac{0}{-60}$$

Divide 0 by -60.

$$x=0$$

Now solve the equation x=\frac{-3±3}{-60} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{-60}$$

Reduce the fraction \frac{-6}{-60} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{10}$$

The equation is now solved.

$$x=0 x=\frac{1}{10}$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x=2\times 15x^{2}$$

Multiply 2 and 15 to get 30.

$$3x=30x^{2}$$

Subtract 30x^{2} from both sides.

$$3x-30x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-30x^{2}+3x=0$$

Divide both sides by -30.

$$\frac{-30x^{2}+3x}{-30}=\frac{0}{-30}$$

Dividing by -30 undoes the multiplication by -30.

$$x^{2}+\frac{3}{-30}x=\frac{0}{-30}$$

Reduce the fraction \frac{3}{-30} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{10}x=\frac{0}{-30}$$

Divide 0 by -30.

$$x^{2}-\frac{1}{10}x=0$$

Divide -\frac{1}{10}, the coefficient of the x term, by 2 to get -\frac{1}{20}. Then add the square of -\frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=\left(-\frac{1}{20}\right)^{2}$$

Square -\frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{1}{400}$$

Factor x^{2}-\frac{1}{10}x+\frac{1}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{20}\right)^{2}=\frac{1}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{20}\right)^{2}}=\sqrt{\frac{1}{400}}$$

Simplify.

$$x-\frac{1}{20}=\frac{1}{20} x-\frac{1}{20}=-\frac{1}{20}$$

Add \frac{1}{20} to both sides of the equation.

$$x=\frac{1}{10} x=0$$