Solve 2x^2+4x=11+x^2

$$2 { x }^{ 2 } +4x = 11+ { x }^{ 2 }$$

Solve for x (complex solution)

$$x=\sqrt{15}-2\approx 1.872983346$$

$$x=-\left(\sqrt{15}+2\right)\approx -5.872983346$$

Steps Using the Quadratic Formula

Subtract 11 from both sides.

$$2x^{2}+4x-11=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}+4x-11-x^{2}=0$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}+4x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$x=\frac{-4±\sqrt{16+44}}{2}$$

Add 16 to 44.

$$x=\frac{-4±\sqrt{60}}{2}$$

Take the square root of 60.

$$x=\frac{-4±2\sqrt{15}}{2}$$

Now solve the equation x=\frac{-4±2\sqrt{15}}{2} when ± is plus. Add -4 to 2\sqrt{15}.

$$x=\frac{2\sqrt{15}-4}{2}$$

Divide -4+2\sqrt{15} by 2.

$$x=\sqrt{15}-2$$

Now solve the equation x=\frac{-4±2\sqrt{15}}{2} when ± is minus. Subtract 2\sqrt{15} from -4.

$$x=\frac{-2\sqrt{15}-4}{2}$$

Divide -4-2\sqrt{15} by 2.

$$x=-\sqrt{15}-2$$

The equation is now solved.

$$x=\sqrt{15}-2 x=-\sqrt{15}-2$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$2x^{2}+4x-x^{2}=11$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}+4x=11$$

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.

$$x^{2}+4x+2^{2}=11+2^{2}$$

Square 2.

$$x^{2}+4x+4=11+4$$

Add 11 to 4.

$$x^{2}+4x+4=15$$

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}.

$$\left(x+2\right)^{2}=15$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{15}$$

Simplify.

$$x+2=\sqrt{15} x+2=-\sqrt{15}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{15}-2 x=-\sqrt{15}-2$$

Solve for x

$$x=\sqrt{15}-2\approx 1.872983346$$

$$x=-\sqrt{15}-2\approx -5.872983346$$

Steps Using the Quadratic Formula

Subtract 11 from both sides.

$$2x^{2}+4x-11=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}+4x-11-x^{2}=0$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}+4x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$x=\frac{-4±\sqrt{16+44}}{2}$$

Add 16 to 44.

$$x=\frac{-4±\sqrt{60}}{2}$$

Take the square root of 60.

$$x=\frac{-4±2\sqrt{15}}{2}$$

Now solve the equation x=\frac{-4±2\sqrt{15}}{2} when ± is plus. Add -4 to 2\sqrt{15}.

$$x=\frac{2\sqrt{15}-4}{2}$$

Divide -4+2\sqrt{15} by 2.

$$x=\sqrt{15}-2$$

Now solve the equation x=\frac{-4±2\sqrt{15}}{2} when ± is minus. Subtract 2\sqrt{15} from -4.

$$x=\frac{-2\sqrt{15}-4}{2}$$

Divide -4-2\sqrt{15} by 2.

$$x=-\sqrt{15}-2$$

The equation is now solved.

$$x=\sqrt{15}-2 x=-\sqrt{15}-2$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$2x^{2}+4x-x^{2}=11$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}+4x=11$$

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.

$$x^{2}+4x+2^{2}=11+2^{2}$$

Square 2.

$$x^{2}+4x+4=11+4$$

Add 11 to 4.

$$x^{2}+4x+4=15$$

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}.

$$\left(x+2\right)^{2}=15$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{15}$$

Simplify.

$$x+2=\sqrt{15} x+2=-\sqrt{15}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{15}-2 x=-\sqrt{15}-2$$