Solve 4x-left(x-2right)left(x+2right)=2left(x-2right)

$$4x- \left( x-2 \right) \left( x+2 \right) = 2 \left( x-2 \right)$$

Solve for x

$$x=-2$$

$$x=4$$

Steps Using the Quadratic Formula

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

To find the opposite of x^{2}-4, find the opposite of each term.

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

Use the distributive property to multiply 2 by x-2.

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

Subtract 2x from both sides.

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

Combine 4x and -2x to get 2x.

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

Add 4 to both sides.

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

Add 4 and 4 to get 8.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4\times 8}}{2\left(-1\right)}$$

Multiply 4 times 8.

$$x=\frac{-2±\sqrt{4+32}}{2\left(-1\right)}$$

Add 4 to 32.

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

Take the square root of 36.

$$x=\frac{-2±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±6}{-2}$$

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

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

Divide 4 by -2.

$$x=-2$$

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

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

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-2 x=4$$

Steps for Completing the Square

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

To find the opposite of x^{2}-4, find the opposite of each term.

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

Use the distributive property to multiply 2 by x-2.

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

Subtract 2x from both sides.

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

Combine 4x and -2x to get 2x.

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

Subtract 4 from both sides.

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

Subtract 4 from -4 to get -8.

$$2x-x^{2}=-8$$

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.

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

Divide both sides by -1.

$$\frac{-x^{2}+2x}{-1}=-\frac{8}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{2}{-1}x=-\frac{8}{-1}$$

Divide 2 by -1.

$$x^{2}-2x=-\frac{8}{-1}$$

Divide -8 by -1.

$$x^{2}-2x=8$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=8+1$$

Add 8 to 1.

$$x^{2}-2x+1=9$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-1=3 x-1=-3$$

Add 1 to both sides of the equation.

$$x=4 x=-2$$