Solve 10a=1+3/a

$$10 a = 1 + \frac { 3 } { a }$$

Solve for a

$$a=-\frac{1}{2}=-0.5$$

$$a=\frac{3}{5}=0.6$$

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 a.

$$10aa=a+3$$

Multiply a and a to get a^{2}.

$$10a^{2}=a+3$$

Subtract a from both sides.

$$10a^{2}-a=3$$

Subtract 3 from both sides.

$$10a^{2}-a-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-1\right)±\sqrt{1-4\times 10\left(-3\right)}}{2\times 10}$$

Multiply -4 times 10.

$$a=\frac{-\left(-1\right)±\sqrt{1-40\left(-3\right)}}{2\times 10}$$

Multiply -40 times -3.

$$a=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 10}$$

Add 1 to 120.

$$a=\frac{-\left(-1\right)±\sqrt{121}}{2\times 10}$$

Take the square root of 121.

$$a=\frac{-\left(-1\right)±11}{2\times 10}$$

The opposite of -1 is 1.

$$a=\frac{1±11}{2\times 10}$$

Multiply 2 times 10.

$$a=\frac{1±11}{20}$$

Now solve the equation a=\frac{1±11}{20} when ± is plus. Add 1 to 11.

$$a=\frac{12}{20}$$

Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.

$$a=\frac{3}{5}$$

Now solve the equation a=\frac{1±11}{20} when ± is minus. Subtract 11 from 1.

$$a=-\frac{10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$a=-\frac{1}{2}$$

The equation is now solved.

$$a=\frac{3}{5} a=-\frac{1}{2}$$

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 a.

$$10aa=a+3$$

Multiply a and a to get a^{2}.

$$10a^{2}=a+3$$

Subtract a from both sides.

$$10a^{2}-a=3$$

Divide both sides by 10.

$$\frac{10a^{2}-a}{10}=\frac{3}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$a^{2}-\frac{1}{10}a=\frac{3}{10}$$

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.

$$a^{2}-\frac{1}{10}a+\left(-\frac{1}{20}\right)^{2}=\frac{3}{10}+\left(-\frac{1}{20}\right)^{2}$$

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

$$a^{2}-\frac{1}{10}a+\frac{1}{400}=\frac{3}{10}+\frac{1}{400}$$

Add \frac{3}{10} to \frac{1}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{1}{10}a+\frac{1}{400}=\frac{121}{400}$$

Factor a^{2}-\frac{1}{10}a+\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}.

$$\left(a-\frac{1}{20}\right)^{2}=\frac{121}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{20}\right)^{2}}=\sqrt{\frac{121}{400}}$$

Simplify.

$$a-\frac{1}{20}=\frac{11}{20} a-\frac{1}{20}=-\frac{11}{20}$$

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

$$a=\frac{3}{5} a=-\frac{1}{2}$$