Solve (2x-4)^2-2x(x-2)=48

$$( 2 x - 4 ) ^ { 2 } - 2 x ( x - 2 ) = 48$$

Solve for x

$$x=-2$$

$$x=8$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16-2x\left(x-2\right)=48$$

Subtract 48 from both sides.

$$4x^{2}-16x+16-2x\left(x-2\right)-48=0$$

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

$$4x^{2}-16x+16-2x^{2}+4x-48=0$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-16x+16+4x-48=0$$

Combine -16x and 4x to get -12x.

$$2x^{2}-12x+16-48=0$$

Subtract 48 from 16 to get -32.

$$2x^{2}-12x-32=0$$

Divide both sides by 2.

$$x^{2}-6x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-16\right)=-16$$

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 -16.

$$1,-16 2,-8 4,-4$$

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -6.

$$a=-8 b=2$$

Rewrite x^{2}-6x-16 as \left(x^{2}-8x\right)+\left(2x-16\right).

$$\left(x^{2}-8x\right)+\left(2x-16\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-8\right)+2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+2\right)$$

To find equation solutions, solve x-8=0 and x+2=0.

$$x=8 x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16-2x\left(x-2\right)=48$$

Subtract 48 from both sides.

$$4x^{2}-16x+16-2x\left(x-2\right)-48=0$$

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

$$4x^{2}-16x+16-2x^{2}+4x-48=0$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-16x+16+4x-48=0$$

Combine -16x and 4x to get -12x.

$$2x^{2}-12x+16-48=0$$

Subtract 48 from 16 to get -32.

$$2x^{2}-12x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\left(-32\right)}}{2\times 2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\left(-32\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\left(-32\right)}}{2\times 2}$$

Multiply -8 times -32.

$$x=\frac{-\left(-12\right)±\sqrt{144+256}}{2\times 2}$$

Add 144 to 256.

$$x=\frac{-\left(-12\right)±\sqrt{400}}{2\times 2}$$

Take the square root of 400.

$$x=\frac{-\left(-12\right)±20}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12±20}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±20}{4}$$

Now solve the equation x=\frac{12±20}{4} when ± is plus. Add 12 to 20.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{12±20}{4} when ± is minus. Subtract 20 from 12.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=8 x=-2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16-2x\left(x-2\right)=48$$

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

$$4x^{2}-16x+16-2x^{2}+4x=48$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-16x+16+4x=48$$

Combine -16x and 4x to get -12x.

$$2x^{2}-12x+16=48$$

Subtract 16 from both sides.

$$2x^{2}-12x=48-16$$

Subtract 16 from 48 to get 32.

$$2x^{2}-12x=32$$

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=\frac{32}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=\frac{32}{2}$$

Divide -12 by 2.

$$x^{2}-6x=\frac{32}{2}$$

Divide 32 by 2.

$$x^{2}-6x=16$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=16+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=16+9$$

Add 16 to 9.

$$x^{2}-6x+9=25$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-3=5 x-3=-5$$

Add 3 to both sides of the equation.

$$x=8 x=-2$$