Solve x(x+3)(x-3)=(x^2+2)(x+2)-x

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

Solve for x

$$x=\frac{\sqrt{17}-5}{2}\approx -0.438447187$$

$$x=\frac{-\sqrt{17}-5}{2}\approx -4.561552813$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply x^{2}+3x by x-3 and combine like terms.

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

Use the distributive property to multiply x^{2}+2 by x+2.

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

Combine 2x and -x to get x.

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

Subtract x^{3} from both sides.

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

Combine x^{3} and -x^{3} to get 0.

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

Subtract 2x^{2} from both sides.

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

Subtract x from both sides.

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

Combine -9x and -x to get -10x.

$$-10x-2x^{2}=4$$

Subtract 4 from both sides.

$$-10x-2x^{2}-4=0$$

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.

$$-2x^{2}-10x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -10 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2\right)\left(-4\right)}}{2\left(-2\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2\right)\left(-4\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-10\right)±\sqrt{100+8\left(-4\right)}}{2\left(-2\right)}$$

Multiply 8 times -4.

$$x=\frac{-\left(-10\right)±\sqrt{100-32}}{2\left(-2\right)}$$

Add 100 to -32.

$$x=\frac{-\left(-10\right)±\sqrt{68}}{2\left(-2\right)}$$

Take the square root of 68.

$$x=\frac{-\left(-10\right)±2\sqrt{17}}{2\left(-2\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{10±2\sqrt{17}}{-4}$$

Now solve the equation x=\frac{10±2\sqrt{17}}{-4} when ± is plus. Add 10 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}+10}{-4}$$

Divide 10+2\sqrt{17} by -4.

$$x=\frac{-\sqrt{17}-5}{2}$$

Now solve the equation x=\frac{10±2\sqrt{17}}{-4} when ± is minus. Subtract 2\sqrt{17} from 10.

$$x=\frac{10-2\sqrt{17}}{-4}$$

Divide 10-2\sqrt{17} by -4.

$$x=\frac{\sqrt{17}-5}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{17}-5}{2} x=\frac{\sqrt{17}-5}{2}$$

Steps for Completing the Square

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

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

Use the distributive property to multiply x^{2}+3x by x-3 and combine like terms.

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

Use the distributive property to multiply x^{2}+2 by x+2.

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

Combine 2x and -x to get x.

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

Subtract x^{3} from both sides.

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

Combine x^{3} and -x^{3} to get 0.

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

Subtract 2x^{2} from both sides.

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

Subtract x from both sides.

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

Combine -9x and -x to get -10x.

$$-10x-2x^{2}=4$$

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.

$$-2x^{2}-10x=4$$

Divide both sides by -2.

$$\frac{-2x^{2}-10x}{-2}=\frac{4}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{10}{-2}\right)x=\frac{4}{-2}$$

Divide -10 by -2.

$$x^{2}+5x=\frac{4}{-2}$$

Divide 4 by -2.

$$x^{2}+5x=-2$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-2+\frac{25}{4}$$

Add -2 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{17}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{17}}{2} x+\frac{5}{2}=-\frac{\sqrt{17}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{17}-5}{2} x=\frac{-\sqrt{17}-5}{2}$$