Solve (x-5)(x+1)=2(x-2)^2=

$$( x - 5 ) ( x + 1 ) = 2 ( x - 2 ) ^ { 2 } =$$

Solve for x (complex solution)

$$x=2-3i$$

$$x=2+3i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-5 by x+1 and combine like terms.

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

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

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

Use the distributive property to multiply 2 by x^{2}-4x+4.

$$x^{2}-4x-5=2x^{2}-8x+8$$

Subtract 2x^{2} from both sides.

$$x^{2}-4x-5-2x^{2}=-8x+8$$

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

$$-x^{2}-4x-5=-8x+8$$

Add 8x to both sides.

$$-x^{2}-4x-5+8x=8$$

Combine -4x and 8x to get 4x.

$$-x^{2}+4x-5=8$$

Subtract 8 from both sides.

$$-x^{2}+4x-5-8=0$$

Subtract 8 from -5 to get -13.

$$-x^{2}+4x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\left(-13\right)}}{2\left(-1\right)}$$

Multiply 4 times -13.

$$x=\frac{-4±\sqrt{16-52}}{2\left(-1\right)}$$

Add 16 to -52.

$$x=\frac{-4±\sqrt{-36}}{2\left(-1\right)}$$

Take the square root of -36.

$$x=\frac{-4±6i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±6i}{-2}$$

Now solve the equation x=\frac{-4±6i}{-2} when ± is plus. Add -4 to 6i.

$$x=\frac{-4+6i}{-2}$$

Divide -4+6i by -2.

$$x=2-3i$$

Now solve the equation x=\frac{-4±6i}{-2} when ± is minus. Subtract 6i from -4.

$$x=\frac{-4-6i}{-2}$$

Divide -4-6i by -2.

$$x=2+3i$$

The equation is now solved.

$$x=2-3i x=2+3i$$

Steps for Completing the Square

Use the distributive property to multiply x-5 by x+1 and combine like terms.

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

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

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

Use the distributive property to multiply 2 by x^{2}-4x+4.

$$x^{2}-4x-5=2x^{2}-8x+8$$

Subtract 2x^{2} from both sides.

$$x^{2}-4x-5-2x^{2}=-8x+8$$

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

$$-x^{2}-4x-5=-8x+8$$

Add 8x to both sides.

$$-x^{2}-4x-5+8x=8$$

Combine -4x and 8x to get 4x.

$$-x^{2}+4x-5=8$$

Add 5 to both sides.

$$-x^{2}+4x=8+5$$

Add 8 and 5 to get 13.

$$-x^{2}+4x=13$$

Divide both sides by -1.

$$\frac{-x^{2}+4x}{-1}=\frac{13}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{4}{-1}x=\frac{13}{-1}$$

Divide 4 by -1.

$$x^{2}-4x=\frac{13}{-1}$$

Divide 13 by -1.

$$x^{2}-4x=-13$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-13+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-13+4$$

Add -13 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=3i x-2=-3i$$

Add 2 to both sides of the equation.

$$x=2+3i x=2-3i$$