Solve 12x^2+3x=0

$$12 { x }^{ 2 } +3x=0$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(12x+3\right)=0$$

To find equation solutions, solve x=0 and 12x+3=0.

$$x=0 x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

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.

$$12x^{2}+3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}}}{2\times 12}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-3±3}{24}$$

Now solve the equation x=\frac{-3±3}{24} when ± is plus. Add -3 to 3.

$$x=\frac{0}{24}$$

Divide 0 by 24.

$$x=0$$

Now solve the equation x=\frac{-3±3}{24} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{24}$$

Reduce the fraction \frac{-6}{24} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=0 x=-\frac{1}{4}$$

Steps for Completing the Square

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.

$$12x^{2}+3x=0$$

Divide both sides by 12.

$$\frac{12x^{2}+3x}{12}=\frac{0}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{3}{12}x=\frac{0}{12}$$

Reduce the fraction \frac{3}{12} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{4}x=\frac{0}{12}$$

Divide 0 by 12.

$$x^{2}+\frac{1}{4}x=0$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{1}{8} x+\frac{1}{8}=-\frac{1}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=0 x=-\frac{1}{4}$$