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

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

Solve for x

$$x=2\sqrt{3}+3\approx 6.464101615$$

$$x=3-2\sqrt{3}\approx -0.464101615$$

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}.

$$-1=-\frac{1}{3}\left(x^{2}-6x+9\right)+3$$

Use the distributive property to multiply -\frac{1}{3} by x^{2}-6x+9.

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

Add -3 and 3 to get 0.

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

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

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

Add 1 to both sides.

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

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

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

Square 2.

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

Multiply -4 times -\frac{1}{3}.

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

Add 4 to \frac{4}{3}.

$$x=\frac{-2±\sqrt{\frac{16}{3}}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of \frac{16}{3}.

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

Multiply 2 times -\frac{1}{3}.

$$x=\frac{-2±\frac{4\sqrt{3}}{3}}{-\frac{2}{3}}$$

Now solve the equation x=\frac{-2±\frac{4\sqrt{3}}{3}}{-\frac{2}{3}} when ± is plus. Add -2 to \frac{4\sqrt{3}}{3}.

$$x=\frac{\frac{4\sqrt{3}}{3}-2}{-\frac{2}{3}}$$

Divide -2+\frac{4\sqrt{3}}{3} by -\frac{2}{3} by multiplying -2+\frac{4\sqrt{3}}{3} by the reciprocal of -\frac{2}{3}.

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

Now solve the equation x=\frac{-2±\frac{4\sqrt{3}}{3}}{-\frac{2}{3}} when ± is minus. Subtract \frac{4\sqrt{3}}{3} from -2.

$$x=\frac{-\frac{4\sqrt{3}}{3}-2}{-\frac{2}{3}}$$

Divide -2-\frac{4\sqrt{3}}{3} by -\frac{2}{3} by multiplying -2-\frac{4\sqrt{3}}{3} by the reciprocal of -\frac{2}{3}.

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

The equation is now solved.

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

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}.

$$-1=-\frac{1}{3}\left(x^{2}-6x+9\right)+3$$

Use the distributive property to multiply -\frac{1}{3} by x^{2}-6x+9.

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

Add -3 and 3 to get 0.

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

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

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

Multiply both sides by -3.

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

Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.

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

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

$$x^{2}-6x=-\frac{1}{-\frac{1}{3}}$$

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

$$x^{2}-6x=3$$

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.

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

Square -3.

$$x^{2}-6x+9=3+9$$

Add 3 to 9.

$$x^{2}-6x+9=12$$

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}.

$$\left(x-3\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{12}$$

Simplify.

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

Add 3 to both sides of the equation.

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