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

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

Solve for x

$$x=1$$

$$x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-3\right), the least common multiple of 2,2x\left(3-x\right).

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

Use the distributive property to multiply x by x-3.

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

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

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

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

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

Subtract 3 from 9 to get 6.

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

To find the opposite of 6-6x+2x^{2}, find the opposite of each term.

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

Subtract -6 from both sides.

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

The opposite of -6 is 6.

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

Subtract 6x from both sides.

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

Combine -3x and -6x to get -9x.

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

Add 2x^{2} to both sides.

$$x^{2}-9x+6+2x^{2}=0$$

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

$$3x^{2}-9x+6=0$$

Divide both sides by 3.

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

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

$$a+b=-3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-2 b=-1$$

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

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

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

$$x\left(x-2\right)-\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x-1\right)$$

To find equation solutions, solve x-2=0 and x-1=0.

$$x=2 x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-3\right), the least common multiple of 2,2x\left(3-x\right).

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

Use the distributive property to multiply x by x-3.

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

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

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

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

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

Subtract 3 from 9 to get 6.

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

To find the opposite of 6-6x+2x^{2}, find the opposite of each term.

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

Subtract -6 from both sides.

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

The opposite of -6 is 6.

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

Subtract 6x from both sides.

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

Combine -3x and -6x to get -9x.

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

Add 2x^{2} to both sides.

$$x^{2}-9x+6+2x^{2}=0$$

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

$$3x^{2}-9x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -9 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 3\times 6}}{2\times 3}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 3\times 6}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-9\right)±\sqrt{81-12\times 6}}{2\times 3}$$

Multiply -12 times 6.

$$x=\frac{-\left(-9\right)±\sqrt{81-72}}{2\times 3}$$

Add 81 to -72.

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

Take the square root of 9.

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

The opposite of -9 is 9.

$$x=\frac{9±3}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{9±3}{6}$$

Now solve the equation x=\frac{9±3}{6} when ± is plus. Add 9 to 3.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{9±3}{6} when ± is minus. Subtract 3 from 9.

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

Divide 6 by 6.

$$x=1$$

The equation is now solved.

$$x=2 x=1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-3\right), the least common multiple of 2,2x\left(3-x\right).

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

Use the distributive property to multiply x by x-3.

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

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

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

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

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

Subtract 3 from 9 to get 6.

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

To find the opposite of 6-6x+2x^{2}, find the opposite of each term.

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

Subtract 6x from both sides.

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

Combine -3x and -6x to get -9x.

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

Add 2x^{2} to both sides.

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

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

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

Divide both sides by 3.

$$\frac{3x^{2}-9x}{3}=-\frac{6}{3}$$

Dividing by 3 undoes the multiplication by 3.

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

Divide -9 by 3.

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

Divide -6 by 3.

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

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

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

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

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

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$