Solve (n-7)^2=25

$$( n - 7 ) ^ { 2 } = 25$$

Solve for n

$$n=12$$

$$n=2$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-7\right)^{2}.

$$n^{2}-14n+49=25$$

Subtract 25 from both sides.

$$n^{2}-14n+49-25=0$$

Subtract 25 from 49 to get 24.

$$n^{2}-14n+24=0$$

To solve the equation, factor n^{2}-14n+24 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.

$$a+b=-14 ab=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -14.

$$a=-12 b=-2$$

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

$$\left(n-12\right)\left(n-2\right)$$

To find equation solutions, solve n-12=0 and n-2=0.

$$n=12 n=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-7\right)^{2}.

$$n^{2}-14n+49=25$$

Subtract 25 from both sides.

$$n^{2}-14n+49-25=0$$

Subtract 25 from 49 to get 24.

$$n^{2}-14n+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn+24. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -14.

$$a=-12 b=-2$$

Rewrite n^{2}-14n+24 as \left(n^{2}-12n\right)+\left(-2n+24\right).

$$\left(n^{2}-12n\right)+\left(-2n+24\right)$$

Factor out n in the first and -2 in the second group.

$$n\left(n-12\right)-2\left(n-12\right)$$

Factor out common term n-12 by using distributive property.

$$\left(n-12\right)\left(n-2\right)$$

To find equation solutions, solve n-12=0 and n-2=0.

$$n=12 n=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-7\right)^{2}.

$$n^{2}-14n+49=25$$

Subtract 25 from both sides.

$$n^{2}-14n+49-25=0$$

Subtract 25 from 49 to get 24.

$$n^{2}-14n+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -14 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 24}}{2}$$

Square -14.

$$n=\frac{-\left(-14\right)±\sqrt{196-4\times 24}}{2}$$

Multiply -4 times 24.

$$n=\frac{-\left(-14\right)±\sqrt{196-96}}{2}$$

Add 196 to -96.

$$n=\frac{-\left(-14\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$n=\frac{-\left(-14\right)±10}{2}$$

The opposite of -14 is 14.

$$n=\frac{14±10}{2}$$

Now solve the equation n=\frac{14±10}{2} when ± is plus. Add 14 to 10.

$$n=\frac{24}{2}$$

Divide 24 by 2.

$$n=12$$

Now solve the equation n=\frac{14±10}{2} when ± is minus. Subtract 10 from 14.

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

Divide 4 by 2.

$$n=2$$

The equation is now solved.

$$n=12 n=2$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(n-7\right)^{2}}=\sqrt{25}$$

Simplify.

$$n-7=5 n-7=-5$$

Add 7 to both sides of the equation.

$$n=12 n=2$$