Solve 4(n^2-9)=3(n-12)

$$4 ( n ^ { 2 } - 9 ) = 3 ( n - 12 )$$

Solve for n

$$n=\frac{3}{4}=0.75$$

$$n=0$$

Steps Using Factoring

Use the distributive property to multiply 4 by n^{2}-9.

$$4n^{2}-36=3\left(n-12\right)$$

Use the distributive property to multiply 3 by n-12.

$$4n^{2}-36=3n-36$$

Subtract 3n from both sides.

$$4n^{2}-36-3n=-36$$

Add 36 to both sides.

$$4n^{2}-36-3n+36=0$$

Add -36 and 36 to get 0.

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

Factor out n.

$$n\left(4n-3\right)=0$$

To find equation solutions, solve n=0 and 4n-3=0.

$$n=0 n=\frac{3}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by n^{2}-9.

$$4n^{2}-36=3\left(n-12\right)$$

Use the distributive property to multiply 3 by n-12.

$$4n^{2}-36=3n-36$$

Subtract 3n from both sides.

$$4n^{2}-36-3n=-36$$

Add 36 to both sides.

$$4n^{2}-36-3n+36=0$$

Add -36 and 36 to get 0.

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

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

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 4}$$

Take the square root of \left(-3\right)^{2}.

$$n=\frac{-\left(-3\right)±3}{2\times 4}$$

The opposite of -3 is 3.

$$n=\frac{3±3}{2\times 4}$$

Multiply 2 times 4.

$$n=\frac{3±3}{8}$$

Now solve the equation n=\frac{3±3}{8} when ± is plus. Add 3 to 3.

$$n=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$n=\frac{3}{4}$$

Now solve the equation n=\frac{3±3}{8} when ± is minus. Subtract 3 from 3.

$$n=\frac{0}{8}$$

Divide 0 by 8.

$$n=0$$

The equation is now solved.

$$n=\frac{3}{4} n=0$$

Steps for Completing the Square

Use the distributive property to multiply 4 by n^{2}-9.

$$4n^{2}-36=3\left(n-12\right)$$

Use the distributive property to multiply 3 by n-12.

$$4n^{2}-36=3n-36$$

Subtract 3n from both sides.

$$4n^{2}-36-3n=-36$$

Add 36 to both sides.

$$4n^{2}-3n=-36+36$$

Add -36 and 36 to get 0.

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

Divide both sides by 4.

$$\frac{4n^{2}-3n}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$n^{2}-\frac{3}{4}n=\frac{0}{4}$$

Divide 0 by 4.

$$n^{2}-\frac{3}{4}n=0$$

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{3}{4}n+\left(-\frac{3}{8}\right)^{2}=\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{3}{4}n+\frac{9}{64}=\frac{9}{64}$$

Factor n^{2}-\frac{3}{4}n+\frac{9}{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}.

$$\left(n-\frac{3}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

$$n-\frac{3}{8}=\frac{3}{8} n-\frac{3}{8}=-\frac{3}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$n=\frac{3}{4} n=0$$