Solve n^2=65-8n

$$n ^ { 2 } = 65 - 8 n$$

Solve for n

$$n=-13$$

$$n=5$$

Steps Using Factoring

Subtract 65 from both sides.

$$n^{2}-65=-8n$$

Add 8n to both sides.

$$n^{2}-65+8n=0$$

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

$$n^{2}+8n-65=0$$

To solve the equation, factor n^{2}+8n-65 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=8 ab=-65$$

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 -65.

$$-1,65 -5,13$$

Calculate the sum for each pair.

$$-1+65=64 -5+13=8$$

The solution is the pair that gives sum 8.

$$a=-5 b=13$$

Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.

$$\left(n-5\right)\left(n+13\right)$$

To find equation solutions, solve n-5=0 and n+13=0.

$$n=5 n=-13$$

Steps Using Factoring By Grouping

Subtract 65 from both sides.

$$n^{2}-65=-8n$$

Add 8n to both sides.

$$n^{2}-65+8n=0$$

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

$$n^{2}+8n-65=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-65. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-65\right)=-65$$

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 -65.

$$-1,65 -5,13$$

Calculate the sum for each pair.

$$-1+65=64 -5+13=8$$

The solution is the pair that gives sum 8.

$$a=-5 b=13$$

Rewrite n^{2}+8n-65 as \left(n^{2}-5n\right)+\left(13n-65\right).

$$\left(n^{2}-5n\right)+\left(13n-65\right)$$

Factor out n in the first and 13 in the second group.

$$n\left(n-5\right)+13\left(n-5\right)$$

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

$$\left(n-5\right)\left(n+13\right)$$

To find equation solutions, solve n-5=0 and n+13=0.

$$n=5 n=-13$$

Steps Using the Quadratic Formula

Subtract 65 from both sides.

$$n^{2}-65=-8n$$

Add 8n to both sides.

$$n^{2}-65+8n=0$$

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.

$$n^{2}+8n-65=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-8±\sqrt{8^{2}-4\left(-65\right)}}{2}$$

Square 8.

$$n=\frac{-8±\sqrt{64-4\left(-65\right)}}{2}$$

Multiply -4 times -65.

$$n=\frac{-8±\sqrt{64+260}}{2}$$

Add 64 to 260.

$$n=\frac{-8±\sqrt{324}}{2}$$

Take the square root of 324.

$$n=\frac{-8±18}{2}$$

Now solve the equation n=\frac{-8±18}{2} when ± is plus. Add -8 to 18.

$$n=\frac{10}{2}$$

Divide 10 by 2.

$$n=5$$

Now solve the equation n=\frac{-8±18}{2} when ± is minus. Subtract 18 from -8.

$$n=-\frac{26}{2}$$

Divide -26 by 2.

$$n=-13$$

The equation is now solved.

$$n=5 n=-13$$

Steps for Completing the Square

Add 8n to both sides.

$$n^{2}+8n=65$$

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

$$n^{2}+8n+4^{2}=65+4^{2}$$

Square 4.

$$n^{2}+8n+16=65+16$$

Add 65 to 16.

$$n^{2}+8n+16=81$$

Factor n^{2}+8n+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+4\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+4\right)^{2}}=\sqrt{81}$$

Simplify.

$$n+4=9 n+4=-9$$

Subtract 4 from both sides of the equation.

$$n=5 n=-13$$