Solve 3n^2=32-4n

$$3 n ^ { 2 } = 32 - 4 n$$

Solve for n

$$n=-4$$

$$n = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Factoring By Grouping

Subtract 32 from both sides.

$$3n^{2}-32=-4n$$

Add 4n to both sides.

$$3n^{2}-32+4n=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3n^{2}+4n-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3n^{2}+an+bn-32. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\left(-32\right)=-96$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Calculate the sum for each pair.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

The solution is the pair that gives sum 4.

$$a=-8 b=12$$

Rewrite 3n^{2}+4n-32 as \left(3n^{2}-8n\right)+\left(12n-32\right).

$$\left(3n^{2}-8n\right)+\left(12n-32\right)$$

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

$$n\left(3n-8\right)+4\left(3n-8\right)$$

Factor out common term 3n-8 by using distributive property.

$$\left(3n-8\right)\left(n+4\right)$$

To find equation solutions, solve 3n-8=0 and n+4=0.

$$n=\frac{8}{3} n=-4$$

Steps Using the Quadratic Formula

Subtract 32 from both sides.

$$3n^{2}-32=-4n$$

Add 4n to both sides.

$$3n^{2}-32+4n=0$$

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.

$$3n^{2}+4n-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 4 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-4±\sqrt{4^{2}-4\times 3\left(-32\right)}}{2\times 3}$$

Square 4.

$$n=\frac{-4±\sqrt{16-4\times 3\left(-32\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-4±\sqrt{16-12\left(-32\right)}}{2\times 3}$$

Multiply -12 times -32.

$$n=\frac{-4±\sqrt{16+384}}{2\times 3}$$

Add 16 to 384.

$$n=\frac{-4±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$n=\frac{-4±20}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-4±20}{6}$$

Now solve the equation n=\frac{-4±20}{6} when ± is plus. Add -4 to 20.

$$n=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation n=\frac{-4±20}{6} when ± is minus. Subtract 20 from -4.

$$n=-\frac{24}{6}$$

Divide -24 by 6.

$$n=-4$$

The equation is now solved.

$$n=\frac{8}{3} n=-4$$

Steps for Completing the Square

Add 4n to both sides.

$$3n^{2}+4n=32$$

Divide both sides by 3.

$$\frac{3n^{2}+4n}{3}=\frac{32}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}+\frac{4}{3}n=\frac{32}{3}$$

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

$$n^{2}+\frac{4}{3}n+\left(\frac{2}{3}\right)^{2}=\frac{32}{3}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{4}{3}n+\frac{4}{9}=\frac{32}{3}+\frac{4}{9}$$

Add \frac{32}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{4}{3}n+\frac{4}{9}=\frac{100}{9}$$

Factor n^{2}+\frac{4}{3}n+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{2}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{2}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$n+\frac{2}{3}=\frac{10}{3} n+\frac{2}{3}=-\frac{10}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

$$n=\frac{8}{3} n=-4$$