Solve 13^2=(x-2)^2+25

$$13 ^ { 2 } = ( x - 2 ) ^ { 2 } + 25$$

Solve for x

$$x=14$$

$$x=-10$$

Steps Using Factoring

Calculate 13 to the power of 2 and get 169.

$$169=\left(x-2\right)^{2}+25$$

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

$$169=x^{2}-4x+4+25$$

Add 4 and 25 to get 29.

$$169=x^{2}-4x+29$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+29=169$$

Subtract 169 from both sides.

$$x^{2}-4x+29-169=0$$

Subtract 169 from 29 to get -140.

$$x^{2}-4x-140=0$$

To solve the equation, factor x^{2}-4x-140 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=-140$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -4.

$$a=-14 b=10$$

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

$$\left(x-14\right)\left(x+10\right)$$

To find equation solutions, solve x-14=0 and x+10=0.

$$x=14 x=-10$$

Steps Using Factoring By Grouping

Calculate 13 to the power of 2 and get 169.

$$169=\left(x-2\right)^{2}+25$$

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

$$169=x^{2}-4x+4+25$$

Add 4 and 25 to get 29.

$$169=x^{2}-4x+29$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+29=169$$

Subtract 169 from both sides.

$$x^{2}-4x+29-169=0$$

Subtract 169 from 29 to get -140.

$$x^{2}-4x-140=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-140. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-140\right)=-140$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -4.

$$a=-14 b=10$$

Rewrite x^{2}-4x-140 as \left(x^{2}-14x\right)+\left(10x-140\right).

$$\left(x^{2}-14x\right)+\left(10x-140\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x-14\right)+10\left(x-14\right)$$

Factor out common term x-14 by using distributive property.

$$\left(x-14\right)\left(x+10\right)$$

To find equation solutions, solve x-14=0 and x+10=0.

$$x=14 x=-10$$

Steps Using the Quadratic Formula

Calculate 13 to the power of 2 and get 169.

$$169=\left(x-2\right)^{2}+25$$

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

$$169=x^{2}-4x+4+25$$

Add 4 and 25 to get 29.

$$169=x^{2}-4x+29$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+29=169$$

Subtract 169 from both sides.

$$x^{2}-4x+29-169=0$$

Subtract 169 from 29 to get -140.

$$x^{2}-4x-140=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -140 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-140\right)}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-140\right)}}{2}$$

Multiply -4 times -140.

$$x=\frac{-\left(-4\right)±\sqrt{16+560}}{2}$$

Add 16 to 560.

$$x=\frac{-\left(-4\right)±\sqrt{576}}{2}$$

Take the square root of 576.

$$x=\frac{-\left(-4\right)±24}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±24}{2}$$

Now solve the equation x=\frac{4±24}{2} when ± is plus. Add 4 to 24.

$$x=\frac{28}{2}$$

Divide 28 by 2.

$$x=14$$

Now solve the equation x=\frac{4±24}{2} when ± is minus. Subtract 24 from 4.

$$x=-\frac{20}{2}$$

Divide -20 by 2.

$$x=-10$$

The equation is now solved.

$$x=14 x=-10$$

Steps for Completing the Square

Calculate 13 to the power of 2 and get 169.

$$169=\left(x-2\right)^{2}+25$$

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

$$169=x^{2}-4x+4+25$$

Add 4 and 25 to get 29.

$$169=x^{2}-4x+29$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+29=169$$

Subtract 29 from both sides.

$$x^{2}-4x=169-29$$

Subtract 29 from 169 to get 140.

$$x^{2}-4x=140$$

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

$$x^{2}-4x+\left(-2\right)^{2}=140+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=140+4$$

Add 140 to 4.

$$x^{2}-4x+4=144$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{144}$$

Simplify.

$$x-2=12 x-2=-12$$

Add 2 to both sides of the equation.

$$x=14 x=-10$$