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

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

Solve for x

$$x=3\sqrt{5}+6\approx 12.708203932$$

$$x=6-3\sqrt{5}\approx -0.708203932$$

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}.

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

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

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

Subtract \frac{1}{3}x^{2} from both sides.

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

$$4x+6-\frac{1}{3}x^{2}=3$$

Subtract 3 from both sides.

$$4x+6-\frac{1}{3}x^{2}-3=0$$

Subtract 3 from 6 to get 3.

$$4x+3-\frac{1}{3}x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-\frac{1}{3}x^{2}+4x+3=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-\frac{1}{3}\right)\times 3}}{2\left(-\frac{1}{3}\right)}$$

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

$$x=\frac{-4±\sqrt{16+\frac{4}{3}\times 3}}{2\left(-\frac{1}{3}\right)}$$

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

$$x=\frac{-4±\sqrt{16+4}}{2\left(-\frac{1}{3}\right)}$$

Add 16 to 4.

$$x=\frac{-4±\sqrt{20}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of 20.

$$x=\frac{-4±2\sqrt{5}}{2\left(-\frac{1}{3}\right)}$$

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

$$x=\frac{-4±2\sqrt{5}}{-\frac{2}{3}}$$

Now solve the equation x=\frac{-4±2\sqrt{5}}{-\frac{2}{3}} when ± is plus. Add -4 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-4}{-\frac{2}{3}}$$

Divide -4+2\sqrt{5} by -\frac{2}{3} by multiplying -4+2\sqrt{5} by the reciprocal of -\frac{2}{3}.

$$x=6-3\sqrt{5}$$

Now solve the equation x=\frac{-4±2\sqrt{5}}{-\frac{2}{3}} when ± is minus. Subtract 2\sqrt{5} from -4.

$$x=\frac{-2\sqrt{5}-4}{-\frac{2}{3}}$$

Divide -4-2\sqrt{5} by -\frac{2}{3} by multiplying -4-2\sqrt{5} by the reciprocal of -\frac{2}{3}.

$$x=3\sqrt{5}+6$$

The equation is now solved.

$$x=6-3\sqrt{5} x=3\sqrt{5}+6$$

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}.

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

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

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

Subtract \frac{1}{3}x^{2} from both sides.

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

Add 2x to both sides.

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

Combine 2x and 2x to get 4x.

$$4x+6-\frac{1}{3}x^{2}=3$$

Subtract 6 from both sides.

$$4x-\frac{1}{3}x^{2}=3-6$$

Subtract 6 from 3 to get -3.

$$4x-\frac{1}{3}x^{2}=-3$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-\frac{1}{3}x^{2}+4x=-3$$

Multiply both sides by -3.

$$\frac{-\frac{1}{3}x^{2}+4x}{-\frac{1}{3}}=-\frac{3}{-\frac{1}{3}}$$

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

$$x^{2}+\frac{4}{-\frac{1}{3}}x=-\frac{3}{-\frac{1}{3}}$$

Divide 4 by -\frac{1}{3} by multiplying 4 by the reciprocal of -\frac{1}{3}.

$$x^{2}-12x=-\frac{3}{-\frac{1}{3}}$$

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

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

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=9+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=9+36$$

Add 9 to 36.

$$x^{2}-12x+36=45$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=45$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{45}$$

Simplify.

$$x-6=3\sqrt{5} x-6=-3\sqrt{5}$$

Add 6 to both sides of the equation.

$$x=3\sqrt{5}+6 x=6-3\sqrt{5}$$