Solve -2x^2+4x+2/(x^2+1)^2=0

$$\frac { - 2 x ^ { 2 } + 4 x + 2 } { ( x ^ { 2 } + 1 ) ^ { 2 } } = 0$$

Solve for x

$$x=\sqrt{2}+1\approx 2.414213562$$

$$x=1-\sqrt{2}\approx -0.414213562$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by \left(x^{2}+1\right)^{2}.

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

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

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

Square 4.

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

Multiply -4 times -2.

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

Multiply 8 times 2.

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

Add 16 to 16.

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

Take the square root of 32.

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

Multiply 2 times -2.

$$x=\frac{-4±4\sqrt{2}}{-4}$$

Now solve the equation x=\frac{-4±4\sqrt{2}}{-4} when ± is plus. Add -4 to 4\sqrt{2}.

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

Divide -4+4\sqrt{2} by -4.

$$x=1-\sqrt{2}$$

Now solve the equation x=\frac{-4±4\sqrt{2}}{-4} when ± is minus. Subtract 4\sqrt{2} from -4.

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

Divide -4-4\sqrt{2} by -4.

$$x=\sqrt{2}+1$$

The equation is now solved.

$$x=1-\sqrt{2} x=\sqrt{2}+1$$

Steps for Completing the Square

Multiply both sides of the equation by \left(x^{2}+1\right)^{2}.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 4 by -2.

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

Divide -2 by -2.

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

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.

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

Add 1 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=\sqrt{2} x-1=-\sqrt{2}$$

Add 1 to both sides of the equation.

$$x=\sqrt{2}+1 x=1-\sqrt{2}$$