Solve 5^2+12^2=(x-5)^2

$$5 ^ { 2 } + 12 ^ { 2 } = ( x - 5 ) ^ { 2 }$$

Solve for x

$$x=18$$

$$x=-8$$

Steps Using Factoring

Calculate 5 to the power of 2 and get 25.

$$25+12^{2}=\left(x-5\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$25+144=\left(x-5\right)^{2}$$

Add 25 and 144 to get 169.

$$169=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$169=x^{2}-10x+25$$

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

$$x^{2}-10x+25=169$$

Subtract 169 from both sides.

$$x^{2}-10x+25-169=0$$

Subtract 169 from 25 to get -144.

$$x^{2}-10x-144=0$$

To solve the equation, factor x^{2}-10x-144 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.

$$a+b=-10 ab=-144$$

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 -144.

$$1,-144 2,-72 3,-48 4,-36 6,-24 8,-18 9,-16 12,-12$$

Calculate the sum for each pair.

$$1-144=-143 2-72=-70 3-48=-45 4-36=-32 6-24=-18 8-18=-10 9-16=-7 12-12=0$$

The solution is the pair that gives sum -10.

$$a=-18 b=8$$

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

$$\left(x-18\right)\left(x+8\right)$$

To find equation solutions, solve x-18=0 and x+8=0.

$$x=18 x=-8$$

Steps Using Factoring By Grouping

Calculate 5 to the power of 2 and get 25.

$$25+12^{2}=\left(x-5\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$25+144=\left(x-5\right)^{2}$$

Add 25 and 144 to get 169.

$$169=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$169=x^{2}-10x+25$$

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

$$x^{2}-10x+25=169$$

Subtract 169 from both sides.

$$x^{2}-10x+25-169=0$$

Subtract 169 from 25 to get -144.

$$x^{2}-10x-144=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-144. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-144\right)=-144$$

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 -144.

$$1,-144 2,-72 3,-48 4,-36 6,-24 8,-18 9,-16 12,-12$$

Calculate the sum for each pair.

$$1-144=-143 2-72=-70 3-48=-45 4-36=-32 6-24=-18 8-18=-10 9-16=-7 12-12=0$$

The solution is the pair that gives sum -10.

$$a=-18 b=8$$

Rewrite x^{2}-10x-144 as \left(x^{2}-18x\right)+\left(8x-144\right).

$$\left(x^{2}-18x\right)+\left(8x-144\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x-18\right)+8\left(x-18\right)$$

Factor out common term x-18 by using distributive property.

$$\left(x-18\right)\left(x+8\right)$$

To find equation solutions, solve x-18=0 and x+8=0.

$$x=18 x=-8$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$25+12^{2}=\left(x-5\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$25+144=\left(x-5\right)^{2}$$

Add 25 and 144 to get 169.

$$169=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$169=x^{2}-10x+25$$

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

$$x^{2}-10x+25=169$$

Subtract 169 from both sides.

$$x^{2}-10x+25-169=0$$

Subtract 169 from 25 to get -144.

$$x^{2}-10x-144=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -144 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-144\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-144\right)}}{2}$$

Multiply -4 times -144.

$$x=\frac{-\left(-10\right)±\sqrt{100+576}}{2}$$

Add 100 to 576.

$$x=\frac{-\left(-10\right)±\sqrt{676}}{2}$$

Take the square root of 676.

$$x=\frac{-\left(-10\right)±26}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±26}{2}$$

Now solve the equation x=\frac{10±26}{2} when ± is plus. Add 10 to 26.

$$x=\frac{36}{2}$$

Divide 36 by 2.

$$x=18$$

Now solve the equation x=\frac{10±26}{2} when ± is minus. Subtract 26 from 10.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=18 x=-8$$

Steps for Completing the Square

Calculate 5 to the power of 2 and get 25.

$$25+12^{2}=\left(x-5\right)^{2}$$

Calculate 12 to the power of 2 and get 144.

$$25+144=\left(x-5\right)^{2}$$

Add 25 and 144 to get 169.

$$169=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$169=x^{2}-10x+25$$

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

$$x^{2}-10x+25=169$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=169$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{169}$$

Simplify.

$$x-5=13 x-5=-13$$

Add 5 to both sides of the equation.

$$x=18 x=-8$$