Solve x^2-2x+x^2-2x-11=8-3

$$x ^ { 2 } - 2 x + x ^ { 2 } - 2 x - 11 = 8 - 3$$

Solve for x

$$x=-2$$

$$x=4$$

Steps Using Factoring By Grouping

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

$$2x^{2}-2x-2x-11=8-3$$

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

$$2x^{2}-4x-11=8-3$$

Subtract 3 from 8 to get 5.

$$2x^{2}-4x-11=5$$

Subtract 5 from both sides.

$$2x^{2}-4x-11-5=0$$

Subtract 5 from -11 to get -16.

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

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-8\right)=-8$$

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 -8.

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

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -2.

$$a=-4 b=2$$

Rewrite x^{2}-2x-8 as \left(x^{2}-4x\right)+\left(2x-8\right).

$$\left(x^{2}-4x\right)+\left(2x-8\right)$$

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

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

Factor out common term x-4 by using distributive property.

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

To find equation solutions, solve x-4=0 and x+2=0.

$$x=4 x=-2$$

Steps Using the Quadratic Formula

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

$$2x^{2}-2x-2x-11=8-3$$

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

$$2x^{2}-4x-11=8-3$$

Subtract 3 from 8 to get 5.

$$2x^{2}-4x-11=5$$

Subtract 5 from both sides.

$$2x^{2}-4x-11-5=0$$

Subtract 5 from -11 to get -16.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

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

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-16\right)}}{2\times 2}$$

Multiply -8 times -16.

$$x=\frac{-\left(-4\right)±\sqrt{16+128}}{2\times 2}$$

Add 16 to 128.

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

Take the square root of 144.

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

The opposite of -4 is 4.

$$x=\frac{4±12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±12}{4}$$

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

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

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

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

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=4 x=-2$$

Steps for Completing the Square

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

$$2x^{2}-2x-2x-11=8-3$$

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

$$2x^{2}-4x-11=8-3$$

Subtract 3 from 8 to get 5.

$$2x^{2}-4x-11=5$$

Add 11 to both sides.

$$2x^{2}-4x=5+11$$

Add 5 and 11 to get 16.

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

Divide both sides by 2.

$$\frac{2x^{2}-4x}{2}=\frac{16}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -4 by 2.

$$x^{2}-2x=\frac{16}{2}$$

Divide 16 by 2.

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

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.

$$x=4 x=-2$$