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

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

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=7$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 3.

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

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

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

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

$$3x^{2}-18x+27-\left(x-1\right)=6x$$

To find the opposite of x-1, find the opposite of each term.

$$3x^{2}-18x+27-x+1=6x$$

Combine -18x and -x to get -19x.

$$3x^{2}-19x+27+1=6x$$

Add 27 and 1 to get 28.

$$3x^{2}-19x+28=6x$$

Subtract 6x from both sides.

$$3x^{2}-19x+28-6x=0$$

Combine -19x and -6x to get -25x.

$$3x^{2}-25x+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+28. To find a and b, set up a system to be solved.

$$a+b=-25 ab=3\times 28=84$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 84.

$$-1,-84 -2,-42 -3,-28 -4,-21 -6,-14 -7,-12$$

Calculate the sum for each pair.

$$-1-84=-85 -2-42=-44 -3-28=-31 -4-21=-25 -6-14=-20 -7-12=-19$$

The solution is the pair that gives sum -25.

$$a=-21 b=-4$$

Rewrite 3x^{2}-25x+28 as \left(3x^{2}-21x\right)+\left(-4x+28\right).

$$\left(3x^{2}-21x\right)+\left(-4x+28\right)$$

Factor out 3x in the first and -4 in the second group.

$$3x\left(x-7\right)-4\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(3x-4\right)$$

To find equation solutions, solve x-7=0 and 3x-4=0.

$$x=7 x=\frac{4}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

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

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

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

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

$$3x^{2}-18x+27-\left(x-1\right)=6x$$

To find the opposite of x-1, find the opposite of each term.

$$3x^{2}-18x+27-x+1=6x$$

Combine -18x and -x to get -19x.

$$3x^{2}-19x+27+1=6x$$

Add 27 and 1 to get 28.

$$3x^{2}-19x+28=6x$$

Subtract 6x from both sides.

$$3x^{2}-19x+28-6x=0$$

Combine -19x and -6x to get -25x.

$$3x^{2}-25x+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -25 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 3\times 28}}{2\times 3}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 3\times 28}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-25\right)±\sqrt{625-12\times 28}}{2\times 3}$$

Multiply -12 times 28.

$$x=\frac{-\left(-25\right)±\sqrt{625-336}}{2\times 3}$$

Add 625 to -336.

$$x=\frac{-\left(-25\right)±\sqrt{289}}{2\times 3}$$

Take the square root of 289.

$$x=\frac{-\left(-25\right)±17}{2\times 3}$$

The opposite of -25 is 25.

$$x=\frac{25±17}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{25±17}{6}$$

Now solve the equation x=\frac{25±17}{6} when ± is plus. Add 25 to 17.

$$x=\frac{42}{6}$$

Divide 42 by 6.

$$x=7$$

Now solve the equation x=\frac{25±17}{6} when ± is minus. Subtract 17 from 25.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

The equation is now solved.

$$x=7 x=\frac{4}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

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

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

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

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

$$3x^{2}-18x+27-\left(x-1\right)=6x$$

To find the opposite of x-1, find the opposite of each term.

$$3x^{2}-18x+27-x+1=6x$$

Combine -18x and -x to get -19x.

$$3x^{2}-19x+27+1=6x$$

Add 27 and 1 to get 28.

$$3x^{2}-19x+28=6x$$

Subtract 6x from both sides.

$$3x^{2}-19x+28-6x=0$$

Combine -19x and -6x to get -25x.

$$3x^{2}-25x+28=0$$

Subtract 28 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-25x=-28$$

Divide both sides by 3.

$$\frac{3x^{2}-25x}{3}=-\frac{28}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{25}{3}x=-\frac{28}{3}$$

Divide -\frac{25}{3}, the coefficient of the x term, by 2 to get -\frac{25}{6}. Then add the square of -\frac{25}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{3}x+\left(-\frac{25}{6}\right)^{2}=-\frac{28}{3}+\left(-\frac{25}{6}\right)^{2}$$

Square -\frac{25}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{3}x+\frac{625}{36}=-\frac{28}{3}+\frac{625}{36}$$

Add -\frac{28}{3} to \frac{625}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{25}{3}x+\frac{625}{36}=\frac{289}{36}$$

Factor x^{2}-\frac{25}{3}x+\frac{625}{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}.

$$\left(x-\frac{25}{6}\right)^{2}=\frac{289}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{6}\right)^{2}}=\sqrt{\frac{289}{36}}$$

Simplify.

$$x-\frac{25}{6}=\frac{17}{6} x-\frac{25}{6}=-\frac{17}{6}$$

Add \frac{25}{6} to both sides of the equation.

$$x=7 x=\frac{4}{3}$$