Solve x^2=(x-2)^2+(x-6/2)^2

$$x ^ { 2 } = ( x - 2 ) ^ { 2 } + ( \frac { x - 6 } { 2 } ) ^ { 2 }$$

Solve for x

$$x=2$$

$$x=26$$

Steps Using the Quadratic Formula

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

$$x^{2}=x^{2}-4x+4+\left(\frac{x-6}{2}\right)^{2}$$

To raise \frac{x-6}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}=x^{2}-4x+4+\frac{\left(x-6\right)^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-4x+4 times \frac{2^{2}}{2^{2}}.

$$x^{2}=\frac{\left(x^{2}-4x+4\right)\times 2^{2}}{2^{2}}+\frac{\left(x-6\right)^{2}}{2^{2}}$$

Since \frac{\left(x^{2}-4x+4\right)\times 2^{2}}{2^{2}} and \frac{\left(x-6\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}=\frac{\left(x^{2}-4x+4\right)\times 2^{2}+\left(x-6\right)^{2}}{2^{2}}$$

Do the multiplications in \left(x^{2}-4x+4\right)\times 2^{2}+\left(x-6\right)^{2}.

$$x^{2}=\frac{4x^{2}-16x+16+x^{2}-12x+36}{2^{2}}$$

Combine like terms in 4x^{2}-16x+16+x^{2}-12x+36.

$$x^{2}=\frac{5x^{2}-28x+52}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}=\frac{5x^{2}-28x+52}{4}$$

Divide each term of 5x^{2}-28x+52 by 4 to get \frac{5}{4}x^{2}-7x+13.

$$x^{2}=\frac{5}{4}x^{2}-7x+13$$

Subtract \frac{5}{4}x^{2} from both sides.

$$x^{2}-\frac{5}{4}x^{2}=-7x+13$$

Combine x^{2} and -\frac{5}{4}x^{2} to get -\frac{1}{4}x^{2}.

$$-\frac{1}{4}x^{2}=-7x+13$$

Add 7x to both sides.

$$-\frac{1}{4}x^{2}+7x=13$$

Subtract 13 from both sides.

$$-\frac{1}{4}x^{2}+7x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, 7 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-\frac{1}{4}\right)\left(-13\right)}}{2\left(-\frac{1}{4}\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-\frac{1}{4}\right)\left(-13\right)}}{2\left(-\frac{1}{4}\right)}$$

Multiply -4 times -\frac{1}{4}.

$$x=\frac{-7±\sqrt{49-13}}{2\left(-\frac{1}{4}\right)}$$

Add 49 to -13.

$$x=\frac{-7±\sqrt{36}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of 36.

$$x=\frac{-7±6}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$x=\frac{-7±6}{-\frac{1}{2}}$$

Now solve the equation x=\frac{-7±6}{-\frac{1}{2}} when ± is plus. Add -7 to 6.

$$x=-\frac{1}{-\frac{1}{2}}$$

Divide -1 by -\frac{1}{2} by multiplying -1 by the reciprocal of -\frac{1}{2}.

$$x=2$$

Now solve the equation x=\frac{-7±6}{-\frac{1}{2}} when ± is minus. Subtract 6 from -7.

$$x=-\frac{13}{-\frac{1}{2}}$$

Divide -13 by -\frac{1}{2} by multiplying -13 by the reciprocal of -\frac{1}{2}.

$$x=26$$

The equation is now solved.

$$x=2 x=26$$

Steps for Completing the Square

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

$$x^{2}=x^{2}-4x+4+\left(\frac{x-6}{2}\right)^{2}$$

To raise \frac{x-6}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}=x^{2}-4x+4+\frac{\left(x-6\right)^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-4x+4 times \frac{2^{2}}{2^{2}}.

$$x^{2}=\frac{\left(x^{2}-4x+4\right)\times 2^{2}}{2^{2}}+\frac{\left(x-6\right)^{2}}{2^{2}}$$

Since \frac{\left(x^{2}-4x+4\right)\times 2^{2}}{2^{2}} and \frac{\left(x-6\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}=\frac{\left(x^{2}-4x+4\right)\times 2^{2}+\left(x-6\right)^{2}}{2^{2}}$$

Do the multiplications in \left(x^{2}-4x+4\right)\times 2^{2}+\left(x-6\right)^{2}.

$$x^{2}=\frac{4x^{2}-16x+16+x^{2}-12x+36}{2^{2}}$$

Combine like terms in 4x^{2}-16x+16+x^{2}-12x+36.

$$x^{2}=\frac{5x^{2}-28x+52}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}=\frac{5x^{2}-28x+52}{4}$$

Divide each term of 5x^{2}-28x+52 by 4 to get \frac{5}{4}x^{2}-7x+13.

$$x^{2}=\frac{5}{4}x^{2}-7x+13$$

Subtract \frac{5}{4}x^{2} from both sides.

$$x^{2}-\frac{5}{4}x^{2}=-7x+13$$

Combine x^{2} and -\frac{5}{4}x^{2} to get -\frac{1}{4}x^{2}.

$$-\frac{1}{4}x^{2}=-7x+13$$

Add 7x to both sides.

$$-\frac{1}{4}x^{2}+7x=13$$

Multiply both sides by -4.

$$\frac{-\frac{1}{4}x^{2}+7x}{-\frac{1}{4}}=\frac{13}{-\frac{1}{4}}$$

Dividing by -\frac{1}{4} undoes the multiplication by -\frac{1}{4}.

$$x^{2}+\frac{7}{-\frac{1}{4}}x=\frac{13}{-\frac{1}{4}}$$

Divide 7 by -\frac{1}{4} by multiplying 7 by the reciprocal of -\frac{1}{4}.

$$x^{2}-28x=\frac{13}{-\frac{1}{4}}$$

Divide 13 by -\frac{1}{4} by multiplying 13 by the reciprocal of -\frac{1}{4}.

$$x^{2}-28x=-52$$

Divide -28, the coefficient of the x term, by 2 to get -14. Then add the square of -14 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-28x+\left(-14\right)^{2}=-52+\left(-14\right)^{2}$$

Square -14.

$$x^{2}-28x+196=-52+196$$

Add -52 to 196.

$$x^{2}-28x+196=144$$

Factor x^{2}-28x+196. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-14\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-14\right)^{2}}=\sqrt{144}$$

Simplify.

$$x-14=12 x-14=-12$$

Add 14 to both sides of the equation.

$$x=26 x=2$$