Solve (x-6)^2+(-x+12)^2=50

$$( x - 6 ) ^ { 2 } + ( - x + 12 ) ^ { 2 } = 50$$

Solve for x

$$x=5$$

$$x=13$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36+\left(-x+12\right)^{2}=50$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+12\right)^{2}.

$$x^{2}-12x+36+\left(-x\right)^{2}+24\left(-x\right)+144=50$$

Calculate -x to the power of 2 and get x^{2}.

$$x^{2}-12x+36+x^{2}+24\left(-x\right)+144=50$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-12x+36+24\left(-x\right)+144=50$$

Add 36 and 144 to get 180.

$$2x^{2}-12x+180+24\left(-x\right)=50$$

Subtract 50 from both sides.

$$2x^{2}-12x+180+24\left(-x\right)-50=0$$

Subtract 50 from 180 to get 130.

$$2x^{2}-12x+130+24\left(-x\right)=0$$

Multiply 24 and -1 to get -24.

$$2x^{2}-12x+130-24x=0$$

Combine -12x and -24x to get -36x.

$$2x^{2}-36x+130=0$$

Divide both sides by 2.

$$x^{2}-18x+65=0$$

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

$$a+b=-18 ab=1\times 65=65$$

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 65.

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

Calculate the sum for each pair.

$$-1-65=-66 -5-13=-18$$

The solution is the pair that gives sum -18.

$$a=-13 b=-5$$

Rewrite x^{2}-18x+65 as \left(x^{2}-13x\right)+\left(-5x+65\right).

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

Factor out x in the first and -5 in the second group.

$$x\left(x-13\right)-5\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(x-5\right)$$

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

$$x=13 x=5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36+\left(-x+12\right)^{2}=50$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+12\right)^{2}.

$$x^{2}-12x+36+\left(-x\right)^{2}+24\left(-x\right)+144=50$$

Calculate -x to the power of 2 and get x^{2}.

$$x^{2}-12x+36+x^{2}+24\left(-x\right)+144=50$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-12x+36+24\left(-x\right)+144=50$$

Add 36 and 144 to get 180.

$$2x^{2}-12x+180+24\left(-x\right)=50$$

Subtract 50 from both sides.

$$2x^{2}-12x+180+24\left(-x\right)-50=0$$

Subtract 50 from 180 to get 130.

$$2x^{2}-12x+130+24\left(-x\right)=0$$

Multiply 24 and -1 to get -24.

$$2x^{2}-12x+130-24x=0$$

Combine -12x and -24x to get -36x.

$$2x^{2}-36x+130=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -36 for b, and 130 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 2\times 130}}{2\times 2}$$

Square -36.

$$x=\frac{-\left(-36\right)±\sqrt{1296-4\times 2\times 130}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-36\right)±\sqrt{1296-8\times 130}}{2\times 2}$$

Multiply -8 times 130.

$$x=\frac{-\left(-36\right)±\sqrt{1296-1040}}{2\times 2}$$

Add 1296 to -1040.

$$x=\frac{-\left(-36\right)±\sqrt{256}}{2\times 2}$$

Take the square root of 256.

$$x=\frac{-\left(-36\right)±16}{2\times 2}$$

The opposite of -36 is 36.

$$x=\frac{36±16}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{36±16}{4}$$

Now solve the equation x=\frac{36±16}{4} when ± is plus. Add 36 to 16.

$$x=\frac{52}{4}$$

Divide 52 by 4.

$$x=13$$

Now solve the equation x=\frac{36±16}{4} when ± is minus. Subtract 16 from 36.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

The equation is now solved.

$$x=13 x=5$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$x^{2}-12x+36+\left(-x+12\right)^{2}=50$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+12\right)^{2}.

$$x^{2}-12x+36+\left(-x\right)^{2}+24\left(-x\right)+144=50$$

Calculate -x to the power of 2 and get x^{2}.

$$x^{2}-12x+36+x^{2}+24\left(-x\right)+144=50$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-12x+36+24\left(-x\right)+144=50$$

Add 36 and 144 to get 180.

$$2x^{2}-12x+180+24\left(-x\right)=50$$

Subtract 180 from both sides.

$$2x^{2}-12x+24\left(-x\right)=50-180$$

Subtract 180 from 50 to get -130.

$$2x^{2}-12x+24\left(-x\right)=-130$$

Multiply 24 and -1 to get -24.

$$2x^{2}-12x-24x=-130$$

Combine -12x and -24x to get -36x.

$$2x^{2}-36x=-130$$

Divide both sides by 2.

$$\frac{2x^{2}-36x}{2}=-\frac{130}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{36}{2}\right)x=-\frac{130}{2}$$

Divide -36 by 2.

$$x^{2}-18x=-\frac{130}{2}$$

Divide -130 by 2.

$$x^{2}-18x=-65$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-18x+\left(-9\right)^{2}=-65+\left(-9\right)^{2}$$

Square -9.

$$x^{2}-18x+81=-65+81$$

Add -65 to 81.

$$x^{2}-18x+81=16$$

Factor x^{2}-18x+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-9\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-9=4 x-9=-4$$

Add 9 to both sides of the equation.

$$x=13 x=5$$