Solve 0=-1/9(x-3)^2+2

$$0 = - \frac { 1 } { 9 } ( x - 3 ) ^ { 2 } + 2$$

Solve for x

$$x=3\sqrt{2}+3\approx 7.242640687$$

$$x=3-3\sqrt{2}\approx -1.242640687$$

Steps Using the Quadratic Formula

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

$$0=-\frac{1}{9}\left(x^{2}-6x+9\right)+2$$

Use the distributive property to multiply -\frac{1}{9} by x^{2}-6x+9.

$$0=-\frac{1}{9}x^{2}+\frac{2}{3}x-1+2$$

Add -1 and 2 to get 1.

$$0=-\frac{1}{9}x^{2}+\frac{2}{3}x+1$$

Swap sides so that all variable terms are on the left hand side.

$$-\frac{1}{9}x^{2}+\frac{2}{3}x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{9} for a, \frac{2}{3} for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{2}{3}±\sqrt{\left(\frac{2}{3}\right)^{2}-4\left(-\frac{1}{9}\right)}}{2\left(-\frac{1}{9}\right)}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-4\left(-\frac{1}{9}\right)}}{2\left(-\frac{1}{9}\right)}$$

Multiply -4 times -\frac{1}{9}.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4+4}{9}}}{2\left(-\frac{1}{9}\right)}$$

Add \frac{4}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{8}{9}}}{2\left(-\frac{1}{9}\right)}$$

Take the square root of \frac{8}{9}.

$$x=\frac{-\frac{2}{3}±\frac{2\sqrt{2}}{3}}{2\left(-\frac{1}{9}\right)}$$

Multiply 2 times -\frac{1}{9}.

$$x=\frac{-\frac{2}{3}±\frac{2\sqrt{2}}{3}}{-\frac{2}{9}}$$

Now solve the equation x=\frac{-\frac{2}{3}±\frac{2\sqrt{2}}{3}}{-\frac{2}{9}} when ± is plus. Add -\frac{2}{3} to \frac{2\sqrt{2}}{3}.

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

Divide \frac{-2+2\sqrt{2}}{3} by -\frac{2}{9} by multiplying \frac{-2+2\sqrt{2}}{3} by the reciprocal of -\frac{2}{9}.

$$x=3-3\sqrt{2}$$

Now solve the equation x=\frac{-\frac{2}{3}±\frac{2\sqrt{2}}{3}}{-\frac{2}{9}} when ± is minus. Subtract \frac{2\sqrt{2}}{3} from -\frac{2}{3}.

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

Divide \frac{-2-2\sqrt{2}}{3} by -\frac{2}{9} by multiplying \frac{-2-2\sqrt{2}}{3} by the reciprocal of -\frac{2}{9}.

$$x=3\sqrt{2}+3$$

The equation is now solved.

$$x=3-3\sqrt{2} x=3\sqrt{2}+3$$

Steps for Completing the Square

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

$$0=-\frac{1}{9}\left(x^{2}-6x+9\right)+2$$

Use the distributive property to multiply -\frac{1}{9} by x^{2}-6x+9.

$$0=-\frac{1}{9}x^{2}+\frac{2}{3}x-1+2$$

Add -1 and 2 to get 1.

$$0=-\frac{1}{9}x^{2}+\frac{2}{3}x+1$$

Swap sides so that all variable terms are on the left hand side.

$$-\frac{1}{9}x^{2}+\frac{2}{3}x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{9}x^{2}+\frac{2}{3}x=-1$$

Multiply both sides by -9.

$$\frac{-\frac{1}{9}x^{2}+\frac{2}{3}x}{-\frac{1}{9}}=-\frac{1}{-\frac{1}{9}}$$

Dividing by -\frac{1}{9} undoes the multiplication by -\frac{1}{9}.

$$x^{2}+\frac{\frac{2}{3}}{-\frac{1}{9}}x=-\frac{1}{-\frac{1}{9}}$$

Divide \frac{2}{3} by -\frac{1}{9} by multiplying \frac{2}{3} by the reciprocal of -\frac{1}{9}.

$$x^{2}-6x=-\frac{1}{-\frac{1}{9}}$$

Divide -1 by -\frac{1}{9} by multiplying -1 by the reciprocal of -\frac{1}{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.

$$x^{2}-6x+9=18$$

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}.

$$\left(x-3\right)^{2}=18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{18}$$

Simplify.

$$x-3=3\sqrt{2} x-3=-3\sqrt{2}$$

Add 3 to both sides of the equation.

$$x=3\sqrt{2}+3 x=3-3\sqrt{2}$$