Solve 2-1quad2x=x^2+2x

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

Solve for x (complex solution)

$$x=\sqrt{6}-2\approx 0.449489743$$

$$x=-\left(\sqrt{6}+2\right)\approx -4.449489743$$

Steps Using the Quadratic Formula

Multiply 1 and 2 to get 2.

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

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

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8}}{2\left(-1\right)}$$

Add 16 to 8.

$$x=\frac{-\left(-4\right)±\sqrt{24}}{2\left(-1\right)}$$

Take the square root of 24.

$$x=\frac{-\left(-4\right)±2\sqrt{6}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{6}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{6}}{-2} when ± is plus. Add 4 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}+4}{-2}$$

Divide 4+2\sqrt{6} by -2.

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

Now solve the equation x=\frac{4±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from 4.

$$x=\frac{4-2\sqrt{6}}{-2}$$

Divide 4-2\sqrt{6} by -2.

$$x=\sqrt{6}-2$$

The equation is now solved.

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

Steps for Completing the Square

Multiply 1 and 2 to get 2.

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

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

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

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

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

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

Divide -2 by -1.

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

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 2.

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

Add 2 to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=\sqrt{6} x+2=-\sqrt{6}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{6}-2 x=-\sqrt{6}-2$$

Solve for x

$$x=\sqrt{6}-2\approx 0.449489743$$

$$x=-\sqrt{6}-2\approx -4.449489743$$

Steps Using the Quadratic Formula

Multiply 1 and 2 to get 2.

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

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

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8}}{2\left(-1\right)}$$

Add 16 to 8.

$$x=\frac{-\left(-4\right)±\sqrt{24}}{2\left(-1\right)}$$

Take the square root of 24.

$$x=\frac{-\left(-4\right)±2\sqrt{6}}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±2\sqrt{6}}{-2}$$

Now solve the equation x=\frac{4±2\sqrt{6}}{-2} when ± is plus. Add 4 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}+4}{-2}$$

Divide 4+2\sqrt{6} by -2.

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

Now solve the equation x=\frac{4±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from 4.

$$x=\frac{4-2\sqrt{6}}{-2}$$

Divide 4-2\sqrt{6} by -2.

$$x=\sqrt{6}-2$$

The equation is now solved.

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

Steps for Completing the Square

Multiply 1 and 2 to get 2.

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

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

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

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

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=-\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

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

Divide -2 by -1.

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

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 2.

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

Add 2 to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=\sqrt{6} x+2=-\sqrt{6}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{6}-2 x=-\sqrt{6}-2$$