Solve a-5a-2/12=a/4a

$$\frac { a - 5 a - 2 } { 12 } = \frac { a } { 4 a }$$

Solve for a

$$a = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

Steps Using Factoring

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12a, the least common multiple of 12,4a.

$$a\left(a-5a-2\right)=3a$$

Combine a and -5a to get -4a.

$$a\left(-4a-2\right)=3a$$

Use the distributive property to multiply a by -4a-2.

$$-4a^{2}-2a=3a$$

Subtract 3a from both sides.

$$-4a^{2}-2a-3a=0$$

Combine -2a and -3a to get -5a.

$$-4a^{2}-5a=0$$

Factor out a.

$$a\left(-4a-5\right)=0$$

To find equation solutions, solve a=0 and -4a-5=0.

$$a=0 a=-\frac{5}{4}$$

Variable a cannot be equal to 0.

$$a=-\frac{5}{4}$$

Steps Using the Quadratic Formula

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12a, the least common multiple of 12,4a.

$$a\left(a-5a-2\right)=3a$$

Combine a and -5a to get -4a.

$$a\left(-4a-2\right)=3a$$

Use the distributive property to multiply a by -4a-2.

$$-4a^{2}-2a=3a$$

Subtract 3a from both sides.

$$-4a^{2}-2a-3a=0$$

Combine -2a and -3a to get -5a.

$$-4a^{2}-5a=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\left(-4\right)}$$

Take the square root of \left(-5\right)^{2}.

$$a=\frac{-\left(-5\right)±5}{2\left(-4\right)}$$

The opposite of -5 is 5.

$$a=\frac{5±5}{2\left(-4\right)}$$

Multiply 2 times -4.

$$a=\frac{5±5}{-8}$$

Now solve the equation a=\frac{5±5}{-8} when ± is plus. Add 5 to 5.

$$a=\frac{10}{-8}$$

Reduce the fraction \frac{10}{-8} to lowest terms by extracting and canceling out 2.

$$a=-\frac{5}{4}$$

Now solve the equation a=\frac{5±5}{-8} when ± is minus. Subtract 5 from 5.

$$a=\frac{0}{-8}$$

Divide 0 by -8.

$$a=0$$

The equation is now solved.

$$a=-\frac{5}{4} a=0$$

Variable a cannot be equal to 0.

$$a=-\frac{5}{4}$$

Steps for Completing the Square

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12a, the least common multiple of 12,4a.

$$a\left(a-5a-2\right)=3a$$

Combine a and -5a to get -4a.

$$a\left(-4a-2\right)=3a$$

Use the distributive property to multiply a by -4a-2.

$$-4a^{2}-2a=3a$$

Subtract 3a from both sides.

$$-4a^{2}-2a-3a=0$$

Combine -2a and -3a to get -5a.

$$-4a^{2}-5a=0$$

Divide both sides by -4.

$$\frac{-4a^{2}-5a}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$a^{2}+\left(-\frac{5}{-4}\right)a=\frac{0}{-4}$$

Divide -5 by -4.

$$a^{2}+\frac{5}{4}a=\frac{0}{-4}$$

Divide 0 by -4.

$$a^{2}+\frac{5}{4}a=0$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{5}{4}a+\left(\frac{5}{8}\right)^{2}=\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{5}{4}a+\frac{25}{64}=\frac{25}{64}$$

Factor a^{2}+\frac{5}{4}a+\frac{25}{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}.

$$\left(a+\frac{5}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{5}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$a+\frac{5}{8}=\frac{5}{8} a+\frac{5}{8}=-\frac{5}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$a=0 a=-\frac{5}{4}$$

Variable a cannot be equal to 0.

$$a=-\frac{5}{4}$$