Solve (x-5)^2+(0-5)^2=169

$$( x - 5 ) ^ { 2 } + ( 0 - 5 ) ^ { 2 } = 169$$

Solve for x

$$x=-7$$

$$x=17$$

Steps Using Factoring

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

$$x^{2}-10x+25+\left(0-5\right)^{2}=169$$

Subtract 5 from 0 to get -5.

$$x^{2}-10x+25+\left(-5\right)^{2}=169$$

Calculate -5 to the power of 2 and get 25.

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

Add 25 and 25 to get 50.

$$x^{2}-10x+50=169$$

Subtract 169 from both sides.

$$x^{2}-10x+50-169=0$$

Subtract 169 from 50 to get -119.

$$x^{2}-10x-119=0$$

To solve the equation, factor x^{2}-10x-119 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.

$$a+b=-10 ab=-119$$

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 -119.

$$1,-119 7,-17$$

Calculate the sum for each pair.

$$1-119=-118 7-17=-10$$

The solution is the pair that gives sum -10.

$$a=-17 b=7$$

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

$$\left(x-17\right)\left(x+7\right)$$

To find equation solutions, solve x-17=0 and x+7=0.

$$x=17 x=-7$$

Steps Using Factoring By Grouping

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

$$x^{2}-10x+25+\left(0-5\right)^{2}=169$$

Subtract 5 from 0 to get -5.

$$x^{2}-10x+25+\left(-5\right)^{2}=169$$

Calculate -5 to the power of 2 and get 25.

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

Add 25 and 25 to get 50.

$$x^{2}-10x+50=169$$

Subtract 169 from both sides.

$$x^{2}-10x+50-169=0$$

Subtract 169 from 50 to get -119.

$$x^{2}-10x-119=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-119. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-119\right)=-119$$

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 -119.

$$1,-119 7,-17$$

Calculate the sum for each pair.

$$1-119=-118 7-17=-10$$

The solution is the pair that gives sum -10.

$$a=-17 b=7$$

Rewrite x^{2}-10x-119 as \left(x^{2}-17x\right)+\left(7x-119\right).

$$\left(x^{2}-17x\right)+\left(7x-119\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-17\right)+7\left(x-17\right)$$

Factor out common term x-17 by using distributive property.

$$\left(x-17\right)\left(x+7\right)$$

To find equation solutions, solve x-17=0 and x+7=0.

$$x=17 x=-7$$

Steps Using the Quadratic Formula

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

$$x^{2}-10x+25+\left(0-5\right)^{2}=169$$

Subtract 5 from 0 to get -5.

$$x^{2}-10x+25+\left(-5\right)^{2}=169$$

Calculate -5 to the power of 2 and get 25.

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

Add 25 and 25 to get 50.

$$x^{2}-10x+50=169$$

Subtract 169 from both sides.

$$x^{2}-10x+50-169=0$$

Subtract 169 from 50 to get -119.

$$x^{2}-10x-119=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -119 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-119\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-119\right)}}{2}$$

Multiply -4 times -119.

$$x=\frac{-\left(-10\right)±\sqrt{100+476}}{2}$$

Add 100 to 476.

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

Take the square root of 576.

$$x=\frac{-\left(-10\right)±24}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±24}{2}$$

Now solve the equation x=\frac{10±24}{2} when ± is plus. Add 10 to 24.

$$x=\frac{34}{2}$$

Divide 34 by 2.

$$x=17$$

Now solve the equation x=\frac{10±24}{2} when ± is minus. Subtract 24 from 10.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=17 x=-7$$

Steps for Completing the Square

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

$$x^{2}-10x+25+\left(0-5\right)^{2}=169$$

Subtract 5 from 0 to get -5.

$$x^{2}-10x+25+\left(-5\right)^{2}=169$$

Calculate -5 to the power of 2 and get 25.

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

Add 25 and 25 to get 50.

$$x^{2}-10x+50=169$$

Subtract 50 from both sides.

$$x^{2}-10x=169-50$$

Subtract 50 from 169 to get 119.

$$x^{2}-10x=119$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=119+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=119+25$$

Add 119 to 25.

$$x^{2}-10x+25=144$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-5=12 x-5=-12$$

Add 5 to both sides of the equation.

$$x=17 x=-7$$