Solve 2x^2-4x+2+3=9

$$2 { x }^{ 2 } -4x+2+3 = 9$$

Solve for x

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=1-\sqrt{3}\approx -0.732050808$$

Steps Using the Quadratic Formula

Add 2 and 3 to get 5.

$$2x^{2}-4x+5=9$$

Subtract 9 from both sides.

$$2x^{2}-4x+5-9=0$$

Subtract 9 from 5 to get -4.

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

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

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

Square -4.

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

Multiply -4 times 2.

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

Multiply -8 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\times 2}$$

Add 16 to 32.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2\times 2}$$

Take the square root of 48.

$$x=\frac{-\left(-4\right)±4\sqrt{3}}{2\times 2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{3}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±4\sqrt{3}}{4}$$

Now solve the equation x=\frac{4±4\sqrt{3}}{4} when ± is plus. Add 4 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+4}{4}$$

Divide 4+4\sqrt{3} by 4.

$$x=\sqrt{3}+1$$

Now solve the equation x=\frac{4±4\sqrt{3}}{4} when ± is minus. Subtract 4\sqrt{3} from 4.

$$x=\frac{4-4\sqrt{3}}{4}$$

Divide 4-4\sqrt{3} by 4.

$$x=1-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

Steps for Completing the Square

Add 2 and 3 to get 5.

$$2x^{2}-4x+5=9$$

Subtract 5 from both sides.

$$2x^{2}-4x=9-5$$

Subtract 5 from 9 to get 4.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -4 by 2.

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

Divide 4 by 2.

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

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.

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

Add 2 to 1.

$$x^{2}-2x+1=3$$

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}.

$$\left(x-1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{3}$$

Simplify.

$$x-1=\sqrt{3} x-1=-\sqrt{3}$$

Add 1 to both sides of the equation.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$