Solve x^2+(6-2x)^2=36

$$x ^ { 2 } + ( 6 - 2 x ) ^ { 2 } = 36$$

Solve for x

$$x = \frac{24}{5} = 4\frac{4}{5} = 4.8$$

$$x=0$$

Steps Using Factoring

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

$$x^{2}+36-24x+4x^{2}=36$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+36-24x=36$$

Subtract 36 from both sides.

$$5x^{2}+36-24x-36=0$$

Subtract 36 from 36 to get 0.

$$5x^{2}-24x=0$$

Factor out x.

$$x\left(5x-24\right)=0$$

To find equation solutions, solve x=0 and 5x-24=0.

$$x=0 x=\frac{24}{5}$$

Steps Using the Quadratic Formula

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

$$x^{2}+36-24x+4x^{2}=36$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+36-24x=36$$

Subtract 36 from both sides.

$$5x^{2}+36-24x-36=0$$

Subtract 36 from 36 to get 0.

$$5x^{2}-24x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -24 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}}}{2\times 5}$$

Take the square root of \left(-24\right)^{2}.

$$x=\frac{-\left(-24\right)±24}{2\times 5}$$

The opposite of -24 is 24.

$$x=\frac{24±24}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{24±24}{10}$$

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

$$x=\frac{48}{10}$$

Reduce the fraction \frac{48}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{24}{5}$$

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

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

The equation is now solved.

$$x=\frac{24}{5} x=0$$

Steps for Completing the Square

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

$$x^{2}+36-24x+4x^{2}=36$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+36-24x=36$$

Subtract 36 from both sides.

$$5x^{2}-24x=36-36$$

Subtract 36 from 36 to get 0.

$$5x^{2}-24x=0$$

Divide both sides by 5.

$$\frac{5x^{2}-24x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{24}{5}x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}-\frac{24}{5}x=0$$

Divide -\frac{24}{5}, the coefficient of the x term, by 2 to get -\frac{12}{5}. Then add the square of -\frac{12}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{24}{5}x+\left(-\frac{12}{5}\right)^{2}=\left(-\frac{12}{5}\right)^{2}$$

Square -\frac{12}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=\frac{144}{25}$$

Factor x^{2}-\frac{24}{5}x+\frac{144}{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}.

$$\left(x-\frac{12}{5}\right)^{2}=\frac{144}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{12}{5}\right)^{2}}=\sqrt{\frac{144}{25}}$$

Simplify.

$$x-\frac{12}{5}=\frac{12}{5} x-\frac{12}{5}=-\frac{12}{5}$$

Add \frac{12}{5} to both sides of the equation.

$$x=\frac{24}{5} x=0$$