Solve 0=x*2x+(2+8x)*-2+4*(x+1)

$$0 = x \times 2x+(2+8x) \times -2+4 \times (x+1)$$

Solve for x

$$x=6$$

$$x=0$$

Steps Using Factoring

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

$$0=x^{2}\times 2+\left(2+8x\right)\left(-2\right)+4\left(x+1\right)$$

Use the distributive property to multiply 2+8x by -2.

$$0=x^{2}\times 2-4-16x+4\left(x+1\right)$$

Use the distributive property to multiply 4 by x+1.

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

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

$$0=x^{2}\times 2-4-12x+4$$

Add -4 and 4 to get 0.

$$0=x^{2}\times 2-12x$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}\times 2-12x=0$$

Factor out x.

$$x\left(2x-12\right)=0$$

To find equation solutions, solve x=0 and 2x-12=0.

$$x=0 x=6$$

Steps Using the Quadratic Formula

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

$$0=x^{2}\times 2+\left(2+8x\right)\left(-2\right)+4\left(x+1\right)$$

Use the distributive property to multiply 2+8x by -2.

$$0=x^{2}\times 2-4-16x+4\left(x+1\right)$$

Use the distributive property to multiply 4 by x+1.

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

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

$$0=x^{2}\times 2-4-12x+4$$

Add -4 and 4 to get 0.

$$0=x^{2}\times 2-12x$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}\times 2-12x=0$$

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.

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

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

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

Take the square root of \left(-12\right)^{2}.

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

The opposite of -12 is 12.

$$x=\frac{12±12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±12}{4}$$

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

$$x=\frac{24}{4}$$

Divide 24 by 4.

$$x=6$$

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

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=6 x=0$$

Steps for Completing the Square

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

$$0=x^{2}\times 2+\left(2+8x\right)\left(-2\right)+4\left(x+1\right)$$

Use the distributive property to multiply 2+8x by -2.

$$0=x^{2}\times 2-4-16x+4\left(x+1\right)$$

Use the distributive property to multiply 4 by x+1.

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

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

$$0=x^{2}\times 2-4-12x+4$$

Add -4 and 4 to get 0.

$$0=x^{2}\times 2-12x$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}\times 2-12x=0$$

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.

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

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=\frac{0}{2}$$

Divide -12 by 2.

$$x^{2}-6x=\frac{0}{2}$$

Divide 0 by 2.

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

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.

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

Square -3.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 3 to both sides of the equation.

$$x=6 x=0$$