Solve 2x^2+3x/3-x=x-x^2/x-3

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

Solve for x

$$x=-4$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3, the least common multiple of 3-x,x-3.

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

To find the opposite of 2x^{2}+3x, find the opposite of each term.

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

Subtract x from both sides.

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

Combine -3x and -x to get -4x.

$$-2x^{2}-4x=-x^{2}$$

Add x^{2} to both sides.

$$-2x^{2}-4x+x^{2}=0$$

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

$$-x^{2}-4x=0$$

Factor out x.

$$x\left(-x-4\right)=0$$

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

$$x=0 x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3, the least common multiple of 3-x,x-3.

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

To find the opposite of 2x^{2}+3x, find the opposite of each term.

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

Subtract x from both sides.

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

Combine -3x and -x to get -4x.

$$-2x^{2}-4x=-x^{2}$$

Add x^{2} to both sides.

$$-2x^{2}-4x+x^{2}=0$$

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

$$-x^{2}-4x=0$$

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

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

Take the square root of \left(-4\right)^{2}.

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

The opposite of -4 is 4.

$$x=\frac{4±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±4}{-2}$$

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

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

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

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

The equation is now solved.

$$x=-4 x=0$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3, the least common multiple of 3-x,x-3.

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

To find the opposite of 2x^{2}+3x, find the opposite of each term.

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

Subtract x from both sides.

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

Combine -3x and -x to get -4x.

$$-2x^{2}-4x=-x^{2}$$

Add x^{2} to both sides.

$$-2x^{2}-4x+x^{2}=0$$

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

$$-x^{2}-4x=0$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{4}{-1}\right)x=\frac{0}{-1}$$

Divide -4 by -1.

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

Divide 0 by -1.

$$x^{2}+4x=0$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=2^{2}$$

Square 2.

$$x^{2}+4x+4=4$$

Factor x^{2}+4x+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}.

$$\left(x+2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+2=2 x+2=-2$$

Subtract 2 from both sides of the equation.

$$x=0 x=-4$$