Solve x^2-6x+9/2-x+x^2=x-(x-2)

$$\frac { x ^ { 2 } - 6 x + 9 } { 2 } - x + x ^ { 2 } = x - ( x - 2 )$$

Solve for x

$$x=1$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

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

Combine -6x and -2x to get -8x.

$$x^{2}-8x+9+2x^{2}=2x-2\left(x-2\right)$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-8x+9=2x-2\left(x-2\right)$$

Use the distributive property to multiply -2 by x-2.

$$3x^{2}-8x+9=2x-2x+4$$

Combine 2x and -2x to get 0.

$$3x^{2}-8x+9=4$$

Subtract 4 from both sides.

$$3x^{2}-8x+9-4=0$$

Subtract 4 from 9 to get 5.

$$3x^{2}-8x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-8 ab=3\times 5=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite 3x^{2}-8x+5 as \left(3x^{2}-5x\right)+\left(-3x+5\right).

$$\left(3x^{2}-5x\right)+\left(-3x+5\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(3x-5\right)-\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(x-1\right)$$

To find equation solutions, solve 3x-5=0 and x-1=0.

$$x=\frac{5}{3} x=1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Combine -6x and -2x to get -8x.

$$x^{2}-8x+9+2x^{2}=2x-2\left(x-2\right)$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-8x+9=2x-2\left(x-2\right)$$

Use the distributive property to multiply -2 by x-2.

$$3x^{2}-8x+9=2x-2x+4$$

Combine 2x and -2x to get 0.

$$3x^{2}-8x+9=4$$

Subtract 4 from both sides.

$$3x^{2}-8x+9-4=0$$

Subtract 4 from 9 to get 5.

$$3x^{2}-8x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -8 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 5}}{2\times 3}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2\times 3}$$

Add 64 to -60.

$$x=\frac{-\left(-8\right)±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

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

The opposite of -8 is 8.

$$x=\frac{8±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±2}{6}$$

Now solve the equation x=\frac{8±2}{6} when ± is plus. Add 8 to 2.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{8±2}{6} when ± is minus. Subtract 2 from 8.

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

Divide 6 by 6.

$$x=1$$

The equation is now solved.

$$x=\frac{5}{3} x=1$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Combine -6x and -2x to get -8x.

$$x^{2}-8x+9+2x^{2}=2x-2\left(x-2\right)$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-8x+9=2x-2\left(x-2\right)$$

Use the distributive property to multiply -2 by x-2.

$$3x^{2}-8x+9=2x-2x+4$$

Combine 2x and -2x to get 0.

$$3x^{2}-8x+9=4$$

Subtract 9 from both sides.

$$3x^{2}-8x=4-9$$

Subtract 9 from 4 to get -5.

$$3x^{2}-8x=-5$$

Divide both sides by 3.

$$\frac{3x^{2}-8x}{3}=-\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{8}{3}x=-\frac{5}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=-\frac{5}{3}+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=-\frac{5}{3}+\frac{16}{9}$$

Add -\frac{5}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{4}{3} to both sides of the equation.

$$x=\frac{5}{3} x=1$$