Solve 9n^2-23n+20=3n^2

$$9 n ^ { 2 } - 23 n + 20 = 3 n ^ { 2 }$$

Solve for n

$$n = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$n = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Subtract 3n^{2} from both sides.

$$9n^{2}-23n+20-3n^{2}=0$$

Combine 9n^{2} and -3n^{2} to get 6n^{2}.

$$6n^{2}-23n+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6n^{2}+an+bn+20. To find a and b, set up a system to be solved.

$$a+b=-23 ab=6\times 20=120$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -23.

$$a=-15 b=-8$$

Rewrite 6n^{2}-23n+20 as \left(6n^{2}-15n\right)+\left(-8n+20\right).

$$\left(6n^{2}-15n\right)+\left(-8n+20\right)$$

Factor out 3n in the first and -4 in the second group.

$$3n\left(2n-5\right)-4\left(2n-5\right)$$

Factor out common term 2n-5 by using distributive property.

$$\left(2n-5\right)\left(3n-4\right)$$

To find equation solutions, solve 2n-5=0 and 3n-4=0.

$$n=\frac{5}{2} n=\frac{4}{3}$$

Steps Using the Quadratic Formula

Subtract 3n^{2} from both sides.

$$9n^{2}-23n+20-3n^{2}=0$$

Combine 9n^{2} and -3n^{2} to get 6n^{2}.

$$6n^{2}-23n+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -23 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 6\times 20}}{2\times 6}$$

Square -23.

$$n=\frac{-\left(-23\right)±\sqrt{529-4\times 6\times 20}}{2\times 6}$$

Multiply -4 times 6.

$$n=\frac{-\left(-23\right)±\sqrt{529-24\times 20}}{2\times 6}$$

Multiply -24 times 20.

$$n=\frac{-\left(-23\right)±\sqrt{529-480}}{2\times 6}$$

Add 529 to -480.

$$n=\frac{-\left(-23\right)±\sqrt{49}}{2\times 6}$$

Take the square root of 49.

$$n=\frac{-\left(-23\right)±7}{2\times 6}$$

The opposite of -23 is 23.

$$n=\frac{23±7}{2\times 6}$$

Multiply 2 times 6.

$$n=\frac{23±7}{12}$$

Now solve the equation n=\frac{23±7}{12} when ± is plus. Add 23 to 7.

$$n=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation n=\frac{23±7}{12} when ± is minus. Subtract 7 from 23.

$$n=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$n=\frac{5}{2} n=\frac{4}{3}$$

Steps for Completing the Square

Subtract 3n^{2} from both sides.

$$9n^{2}-23n+20-3n^{2}=0$$

Combine 9n^{2} and -3n^{2} to get 6n^{2}.

$$6n^{2}-23n+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$6n^{2}-23n=-20$$

Divide both sides by 6.

$$\frac{6n^{2}-23n}{6}=-\frac{20}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$n^{2}-\frac{23}{6}n=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$n^{2}-\frac{23}{6}n=-\frac{10}{3}$$

Divide -\frac{23}{6}, the coefficient of the x term, by 2 to get -\frac{23}{12}. Then add the square of -\frac{23}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{23}{6}n+\left(-\frac{23}{12}\right)^{2}=-\frac{10}{3}+\left(-\frac{23}{12}\right)^{2}$$

Square -\frac{23}{12} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{23}{6}n+\frac{529}{144}=-\frac{10}{3}+\frac{529}{144}$$

Add -\frac{10}{3} to \frac{529}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{23}{6}n+\frac{529}{144}=\frac{49}{144}$$

Factor n^{2}-\frac{23}{6}n+\frac{529}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{23}{12}\right)^{2}=\frac{49}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{23}{12}\right)^{2}}=\sqrt{\frac{49}{144}}$$

Simplify.

$$n-\frac{23}{12}=\frac{7}{12} n-\frac{23}{12}=-\frac{7}{12}$$

Add \frac{23}{12} to both sides of the equation.

$$n=\frac{5}{2} n=\frac{4}{3}$$