Solve (x-2)^2+[(-1/3x+6)-2]^2=10

$$( x - 2 ) ^ { 2 } + [ ( - \frac { 1 } { 3 } x + 6 ) - 2 ] ^ { 2 } = 10$$

Solve for x

$$x=3$$

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}.

$$x^{2}-4x+4+\left(-\frac{1}{3}x+6-2\right)^{2}=10$$

Subtract 2 from 6 to get 4.

$$x^{2}-4x+4+\left(-\frac{1}{3}x+4\right)^{2}=10$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{1}{3}x+4\right)^{2}.

$$x^{2}-4x+4+\frac{1}{9}x^{2}-\frac{8}{3}x+16=10$$

Combine x^{2} and \frac{1}{9}x^{2} to get \frac{10}{9}x^{2}.

$$\frac{10}{9}x^{2}-4x+4-\frac{8}{3}x+16=10$$

Combine -4x and -\frac{8}{3}x to get -\frac{20}{3}x.

$$\frac{10}{9}x^{2}-\frac{20}{3}x+4+16=10$$

Add 4 and 16 to get 20.

$$\frac{10}{9}x^{2}-\frac{20}{3}x+20=10$$

Subtract 10 from both sides.

$$\frac{10}{9}x^{2}-\frac{20}{3}x+20-10=0$$

Subtract 10 from 20 to get 10.

$$\frac{10}{9}x^{2}-\frac{20}{3}x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{10}{9} for a, -\frac{20}{3} for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\left(-\frac{20}{3}\right)^{2}-4\times \frac{10}{9}\times 10}}{2\times \frac{10}{9}}$$

Square -\frac{20}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\frac{400}{9}-4\times \frac{10}{9}\times 10}}{2\times \frac{10}{9}}$$

Multiply -4 times \frac{10}{9}.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\frac{400}{9}-\frac{40}{9}\times 10}}{2\times \frac{10}{9}}$$

Multiply -\frac{40}{9} times 10.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\frac{400-400}{9}}}{2\times \frac{10}{9}}$$

Add \frac{400}{9} to -\frac{400}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{0}}{2\times \frac{10}{9}}$$

Take the square root of 0.

$$x=-\frac{-\frac{20}{3}}{2\times \frac{10}{9}}$$

The opposite of -\frac{20}{3} is \frac{20}{3}.

$$x=\frac{\frac{20}{3}}{2\times \frac{10}{9}}$$

Multiply 2 times \frac{10}{9}.

$$x=\frac{\frac{20}{3}}{\frac{20}{9}}$$

Divide \frac{20}{3} by \frac{20}{9} by multiplying \frac{20}{3} by the reciprocal of \frac{20}{9}.

$$x=3$$

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}.

$$x^{2}-4x+4+\left(-\frac{1}{3}x+6-2\right)^{2}=10$$

Subtract 2 from 6 to get 4.

$$x^{2}-4x+4+\left(-\frac{1}{3}x+4\right)^{2}=10$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{1}{3}x+4\right)^{2}.

$$x^{2}-4x+4+\frac{1}{9}x^{2}-\frac{8}{3}x+16=10$$

Combine x^{2} and \frac{1}{9}x^{2} to get \frac{10}{9}x^{2}.

$$\frac{10}{9}x^{2}-4x+4-\frac{8}{3}x+16=10$$

Combine -4x and -\frac{8}{3}x to get -\frac{20}{3}x.

$$\frac{10}{9}x^{2}-\frac{20}{3}x+4+16=10$$

Add 4 and 16 to get 20.

$$\frac{10}{9}x^{2}-\frac{20}{3}x+20=10$$

Subtract 20 from both sides.

$$\frac{10}{9}x^{2}-\frac{20}{3}x=10-20$$

Subtract 20 from 10 to get -10.

$$\frac{10}{9}x^{2}-\frac{20}{3}x=-10$$

Divide both sides of the equation by \frac{10}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{10}{9}x^{2}-\frac{20}{3}x}{\frac{10}{9}}=-\frac{10}{\frac{10}{9}}$$

Dividing by \frac{10}{9} undoes the multiplication by \frac{10}{9}.

$$x^{2}+\left(-\frac{\frac{20}{3}}{\frac{10}{9}}\right)x=-\frac{10}{\frac{10}{9}}$$

Divide -\frac{20}{3} by \frac{10}{9} by multiplying -\frac{20}{3} by the reciprocal of \frac{10}{9}.

$$x^{2}-6x=-\frac{10}{\frac{10}{9}}$$

Divide -10 by \frac{10}{9} by multiplying -10 by the reciprocal of \frac{10}{9}.

$$x^{2}-6x=-9$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-9+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-9+9$$

Add -9 to 9.

$$x^{2}-6x+9=0$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-3=0 x-3=0$$

Add 3 to both sides of the equation.

$$x=3 x=3$$

The equation is now solved. Solutions are the same.

$$x=3$$