Solve 2(x-2)+2x^2-6x=16

$$2(x-2)+2 { x }^{ 2 } -6x=16$$

Solve for x

$$x=\sqrt{11}+1\approx 4.31662479$$

$$x=1-\sqrt{11}\approx -2.31662479$$

Steps Using the Quadratic Formula

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

$$2x-4+2x^{2}-6x=16$$

Combine 2x and -6x to get -4x.

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

Subtract 16 from both sides.

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

Subtract 16 from -4 to get -20.

$$-4x-20+2x^{2}=0$$

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.

$$2x^{2}-4x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-20\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-20\right)}}{2\times 2}$$

Multiply -8 times -20.

$$x=\frac{-\left(-4\right)±\sqrt{16+160}}{2\times 2}$$

Add 16 to 160.

$$x=\frac{-\left(-4\right)±\sqrt{176}}{2\times 2}$$

Take the square root of 176.

$$x=\frac{-\left(-4\right)±4\sqrt{11}}{2\times 2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{11}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±4\sqrt{11}}{4}$$

Now solve the equation x=\frac{4±4\sqrt{11}}{4} when ± is plus. Add 4 to 4\sqrt{11}.

$$x=\frac{4\sqrt{11}+4}{4}$$

Divide 4+4\sqrt{11} by 4.

$$x=\sqrt{11}+1$$

Now solve the equation x=\frac{4±4\sqrt{11}}{4} when ± is minus. Subtract 4\sqrt{11} from 4.

$$x=\frac{4-4\sqrt{11}}{4}$$

Divide 4-4\sqrt{11} by 4.

$$x=1-\sqrt{11}$$

The equation is now solved.

$$x=\sqrt{11}+1 x=1-\sqrt{11}$$

Steps for Completing the Square

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

$$2x-4+2x^{2}-6x=16$$

Combine 2x and -6x to get -4x.

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

Add 4 to both sides.

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

Add 16 and 4 to get 20.

$$-4x+2x^{2}=20$$

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.

$$2x^{2}-4x=20$$

Divide both sides by 2.

$$\frac{2x^{2}-4x}{2}=\frac{20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{4}{2}\right)x=\frac{20}{2}$$

Divide -4 by 2.

$$x^{2}-2x=\frac{20}{2}$$

Divide 20 by 2.

$$x^{2}-2x=10$$

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.

$$x^{2}-2x+1=10+1$$

Add 10 to 1.

$$x^{2}-2x+1=11$$

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}.

$$\left(x-1\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{11}$$

Simplify.

$$x-1=\sqrt{11} x-1=-\sqrt{11}$$

Add 1 to both sides of the equation.

$$x=\sqrt{11}+1 x=1-\sqrt{11}$$