Solve 9n^2=4+7n

$$9 n ^ { 2 } = 4 + 7 n$$

Solve for n

$$n = \frac{\sqrt{193} + 7}{18} \approx 1.160691333$$

$$n=\frac{7-\sqrt{193}}{18}\approx -0.382913555$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$9n^{2}-4=7n$$

Subtract 7n from both sides.

$$9n^{2}-4-7n=0$$

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.

$$9n^{2}-7n-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 9\left(-4\right)}}{2\times 9}$$

Square -7.

$$n=\frac{-\left(-7\right)±\sqrt{49-4\times 9\left(-4\right)}}{2\times 9}$$

Multiply -4 times 9.

$$n=\frac{-\left(-7\right)±\sqrt{49-36\left(-4\right)}}{2\times 9}$$

Multiply -36 times -4.

$$n=\frac{-\left(-7\right)±\sqrt{49+144}}{2\times 9}$$

Add 49 to 144.

$$n=\frac{-\left(-7\right)±\sqrt{193}}{2\times 9}$$

The opposite of -7 is 7.

$$n=\frac{7±\sqrt{193}}{2\times 9}$$

Multiply 2 times 9.

$$n=\frac{7±\sqrt{193}}{18}$$

Now solve the equation n=\frac{7±\sqrt{193}}{18} when ± is plus. Add 7 to \sqrt{193}.

$$n=\frac{\sqrt{193}+7}{18}$$

Now solve the equation n=\frac{7±\sqrt{193}}{18} when ± is minus. Subtract \sqrt{193} from 7.

$$n=\frac{7-\sqrt{193}}{18}$$

The equation is now solved.

$$n=\frac{\sqrt{193}+7}{18} n=\frac{7-\sqrt{193}}{18}$$

Steps for Completing the Square

Subtract 7n from both sides.

$$9n^{2}-7n=4$$

Divide both sides by 9.

$$\frac{9n^{2}-7n}{9}=\frac{4}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$n^{2}-\frac{7}{9}n=\frac{4}{9}$$

Divide -\frac{7}{9}, the coefficient of the x term, by 2 to get -\frac{7}{18}. Then add the square of -\frac{7}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{7}{9}n+\left(-\frac{7}{18}\right)^{2}=\frac{4}{9}+\left(-\frac{7}{18}\right)^{2}$$

Square -\frac{7}{18} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{7}{9}n+\frac{49}{324}=\frac{4}{9}+\frac{49}{324}$$

Add \frac{4}{9} to \frac{49}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{7}{9}n+\frac{49}{324}=\frac{193}{324}$$

Factor n^{2}-\frac{7}{9}n+\frac{49}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{7}{18}\right)^{2}=\frac{193}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{7}{18}\right)^{2}}=\sqrt{\frac{193}{324}}$$

Simplify.

$$n-\frac{7}{18}=\frac{\sqrt{193}}{18} n-\frac{7}{18}=-\frac{\sqrt{193}}{18}$$

Add \frac{7}{18} to both sides of the equation.

$$n=\frac{\sqrt{193}+7}{18} n=\frac{7-\sqrt{193}}{18}$$