Solve (x-2)^2-5x=2(x+3)(x-3)

$$( x - 2 ) ^ { 2 } - 5 x = 2 ( x + 3 ) ( x - 3 )$$

Solve for x

$$x=-11$$

$$x=2$$

Steps Using Factoring By Grouping

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

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

Combine -4x and -5x to get -9x.

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

Use the distributive property to multiply 2 by x+3.

$$x^{2}-9x+4=\left(2x+6\right)\left(x-3\right)$$

Use the distributive property to multiply 2x+6 by x-3 and combine like terms.

$$x^{2}-9x+4=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$x^{2}-9x+4-2x^{2}=-18$$

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

$$-x^{2}-9x+4=-18$$

Add 18 to both sides.

$$-x^{2}-9x+4+18=0$$

Add 4 and 18 to get 22.

$$-x^{2}-9x+22=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+22. To find a and b, set up a system to be solved.

$$a+b=-9 ab=-22=-22$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -22.

$$1,-22 2,-11$$

Calculate the sum for each pair.

$$1-22=-21 2-11=-9$$

The solution is the pair that gives sum -9.

$$a=2 b=-11$$

Rewrite -x^{2}-9x+22 as \left(-x^{2}+2x\right)+\left(-11x+22\right).

$$\left(-x^{2}+2x\right)+\left(-11x+22\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(-x+2\right)+11\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(x+11\right)$$

To find equation solutions, solve -x+2=0 and x+11=0.

$$x=2 x=-11$$

Steps Using the Quadratic Formula

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

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

Combine -4x and -5x to get -9x.

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

Use the distributive property to multiply 2 by x+3.

$$x^{2}-9x+4=\left(2x+6\right)\left(x-3\right)$$

Use the distributive property to multiply 2x+6 by x-3 and combine like terms.

$$x^{2}-9x+4=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$x^{2}-9x+4-2x^{2}=-18$$

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

$$-x^{2}-9x+4=-18$$

Add 18 to both sides.

$$-x^{2}-9x+4+18=0$$

Add 4 and 18 to get 22.

$$-x^{2}-9x+22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -9 for b, and 22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-1\right)\times 22}}{2\left(-1\right)}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-1\right)\times 22}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-9\right)±\sqrt{81+4\times 22}}{2\left(-1\right)}$$

Multiply 4 times 22.

$$x=\frac{-\left(-9\right)±\sqrt{81+88}}{2\left(-1\right)}$$

Add 81 to 88.

$$x=\frac{-\left(-9\right)±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-9\right)±13}{2\left(-1\right)}$$

The opposite of -9 is 9.

$$x=\frac{9±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{9±13}{-2}$$

Now solve the equation x=\frac{9±13}{-2} when ± is plus. Add 9 to 13.

$$x=\frac{22}{-2}$$

Divide 22 by -2.

$$x=-11$$

Now solve the equation x=\frac{9±13}{-2} when ± is minus. Subtract 13 from 9.

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

Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=-11 x=2$$

Steps for Completing the Square

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

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

Combine -4x and -5x to get -9x.

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

Use the distributive property to multiply 2 by x+3.

$$x^{2}-9x+4=\left(2x+6\right)\left(x-3\right)$$

Use the distributive property to multiply 2x+6 by x-3 and combine like terms.

$$x^{2}-9x+4=2x^{2}-18$$

Subtract 2x^{2} from both sides.

$$x^{2}-9x+4-2x^{2}=-18$$

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

$$-x^{2}-9x+4=-18$$

Subtract 4 from both sides.

$$-x^{2}-9x=-18-4$$

Subtract 4 from -18 to get -22.

$$-x^{2}-9x=-22$$

Divide both sides by -1.

$$\frac{-x^{2}-9x}{-1}=-\frac{22}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{9}{-1}\right)x=-\frac{22}{-1}$$

Divide -9 by -1.

$$x^{2}+9x=-\frac{22}{-1}$$

Divide -22 by -1.

$$x^{2}+9x=22$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=22+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=22+\frac{81}{4}$$

Add 22 to \frac{81}{4}.

$$x^{2}+9x+\frac{81}{4}=\frac{169}{4}$$

Factor x^{2}+9x+\frac{81}{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}.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{13}{2} x+\frac{9}{2}=-\frac{13}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$x=2 x=-11$$