Solve 3/4*3*2=14/2(n-3)(n-2)

$$\frac { 3 } { 4 \times 3 \times 2 } = \frac { 14 } { 2 ( n - 3 ) ( n - 2 ) }$$

Solve for n

$$n=-5$$

$$n=10$$

Steps Using the Quadratic Formula

Variable n cannot be equal to any of the values 2,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(n-3\right)\left(n-2\right).

$$\frac{1}{12}\left(n-3\right)\left(n-2\right)\times 3=14$$

Use the distributive property to multiply \frac{1}{12} by n-3.

$$\left(\frac{1}{12}n-\frac{1}{4}\right)\left(n-2\right)\times 3=14$$

Use the distributive property to multiply \frac{1}{12}n-\frac{1}{4} by n-2 and combine like terms.

$$\left(\frac{1}{12}n^{2}-\frac{5}{12}n+\frac{1}{2}\right)\times 3=14$$

Use the distributive property to multiply \frac{1}{12}n^{2}-\frac{5}{12}n+\frac{1}{2} by 3.

$$\frac{1}{4}n^{2}-\frac{5}{4}n+\frac{3}{2}=14$$

Subtract 14 from both sides.

$$\frac{1}{4}n^{2}-\frac{5}{4}n+\frac{3}{2}-14=0$$

Subtract 14 from \frac{3}{2} to get -\frac{25}{2}.

$$\frac{1}{4}n^{2}-\frac{5}{4}n-\frac{25}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -\frac{5}{4} for b, and -\frac{25}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\left(-\frac{5}{4}\right)^{2}-4\times \frac{1}{4}\left(-\frac{25}{2}\right)}}{2\times \frac{1}{4}}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$n=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\frac{25}{16}-4\times \frac{1}{4}\left(-\frac{25}{2}\right)}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$n=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\frac{25}{16}-\left(-\frac{25}{2}\right)}}{2\times \frac{1}{4}}$$

Multiply -1 times -\frac{25}{2}.

$$n=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\frac{25}{16}+\frac{25}{2}}}{2\times \frac{1}{4}}$$

Add \frac{25}{16} to \frac{25}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\frac{225}{16}}}{2\times \frac{1}{4}}$$

Take the square root of \frac{225}{16}.

$$n=\frac{-\left(-\frac{5}{4}\right)±\frac{15}{4}}{2\times \frac{1}{4}}$$

The opposite of -\frac{5}{4} is \frac{5}{4}.

$$n=\frac{\frac{5}{4}±\frac{15}{4}}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$n=\frac{\frac{5}{4}±\frac{15}{4}}{\frac{1}{2}}$$

Now solve the equation n=\frac{\frac{5}{4}±\frac{15}{4}}{\frac{1}{2}} when ± is plus. Add \frac{5}{4} to \frac{15}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n=\frac{5}{\frac{1}{2}}$$

Divide 5 by \frac{1}{2} by multiplying 5 by the reciprocal of \frac{1}{2}.

$$n=10$$

Now solve the equation n=\frac{\frac{5}{4}±\frac{15}{4}}{\frac{1}{2}} when ± is minus. Subtract \frac{15}{4} from \frac{5}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$n=-\frac{\frac{5}{2}}{\frac{1}{2}}$$

Divide -\frac{5}{2} by \frac{1}{2} by multiplying -\frac{5}{2} by the reciprocal of \frac{1}{2}.

$$n=-5$$

The equation is now solved.

$$n=10 n=-5$$

Steps for Completing the Square

Variable n cannot be equal to any of the values 2,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(n-3\right)\left(n-2\right).

$$\frac{1}{12}\left(n-3\right)\left(n-2\right)\times 3=14$$

Use the distributive property to multiply \frac{1}{12} by n-3.

$$\left(\frac{1}{12}n-\frac{1}{4}\right)\left(n-2\right)\times 3=14$$

Use the distributive property to multiply \frac{1}{12}n-\frac{1}{4} by n-2 and combine like terms.

$$\left(\frac{1}{12}n^{2}-\frac{5}{12}n+\frac{1}{2}\right)\times 3=14$$

Use the distributive property to multiply \frac{1}{12}n^{2}-\frac{5}{12}n+\frac{1}{2} by 3.

$$\frac{1}{4}n^{2}-\frac{5}{4}n+\frac{3}{2}=14$$

Subtract \frac{3}{2} from both sides.

$$\frac{1}{4}n^{2}-\frac{5}{4}n=14-\frac{3}{2}$$

Subtract \frac{3}{2} from 14 to get \frac{25}{2}.

$$\frac{1}{4}n^{2}-\frac{5}{4}n=\frac{25}{2}$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}n^{2}-\frac{5}{4}n}{\frac{1}{4}}=\frac{\frac{25}{2}}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$n^{2}+\left(-\frac{\frac{5}{4}}{\frac{1}{4}}\right)n=\frac{\frac{25}{2}}{\frac{1}{4}}$$

Divide -\frac{5}{4} by \frac{1}{4} by multiplying -\frac{5}{4} by the reciprocal of \frac{1}{4}.

$$n^{2}-5n=\frac{\frac{25}{2}}{\frac{1}{4}}$$

Divide \frac{25}{2} by \frac{1}{4} by multiplying \frac{25}{2} by the reciprocal of \frac{1}{4}.

$$n^{2}-5n=50$$

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

$$n^{2}-5n+\left(-\frac{5}{2}\right)^{2}=50+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-5n+\frac{25}{4}=50+\frac{25}{4}$$

Add 50 to \frac{25}{4}.

$$n^{2}-5n+\frac{25}{4}=\frac{225}{4}$$

Factor n^{2}-5n+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{5}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{5}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$n-\frac{5}{2}=\frac{15}{2} n-\frac{5}{2}=-\frac{15}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$n=10 n=-5$$