Solve x^2+4x+4=3

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

Solve for x (complex solution)

$$x=\sqrt{3}-2\approx -0.267949192$$

$$x=-\left(\sqrt{3}+2\right)\approx -3.732050808$$

Steps Using the Quadratic Formula

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.

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

Subtract 3 from both sides of the equation.

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

Subtracting 3 from itself leaves 0.

$$x^{2}+4x+4-3=0$$

Subtract 3 from 4.

$$x^{2}+4x+1=0$$

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

$$x=\frac{-4±\sqrt{4^{2}-4}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4}}{2}$$

Add 16 to -4.

$$x=\frac{-4±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-4±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{-4±2\sqrt{3}}{2} when ± is plus. Add -4 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-4}{2}$$

Divide -4+2\sqrt{3} by 2.

$$x=\sqrt{3}-2$$

Now solve the equation x=\frac{-4±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from -4.

$$x=\frac{-2\sqrt{3}-4}{2}$$

Divide -4-2\sqrt{3} by 2.

$$x=-\sqrt{3}-2$$

The equation is now solved.

$$x=\sqrt{3}-2 x=-\sqrt{3}-2$$

Steps for Completing the Square

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=\sqrt{3} x+2=-\sqrt{3}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{3}-2 x=-\sqrt{3}-2$$

Solve for x

$$x=\sqrt{3}-2\approx -0.267949192$$

$$x=-\sqrt{3}-2\approx -3.732050808$$

Steps Using the Quadratic Formula

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.

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

Subtract 3 from both sides of the equation.

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

Subtracting 3 from itself leaves 0.

$$x^{2}+4x+4-3=0$$

Subtract 3 from 4.

$$x^{2}+4x+1=0$$

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

$$x=\frac{-4±\sqrt{4^{2}-4}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4}}{2}$$

Add 16 to -4.

$$x=\frac{-4±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-4±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{-4±2\sqrt{3}}{2} when ± is plus. Add -4 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-4}{2}$$

Divide -4+2\sqrt{3} by 2.

$$x=\sqrt{3}-2$$

Now solve the equation x=\frac{-4±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from -4.

$$x=\frac{-2\sqrt{3}-4}{2}$$

Divide -4-2\sqrt{3} by 2.

$$x=-\sqrt{3}-2$$

The equation is now solved.

$$x=\sqrt{3}-2 x=-\sqrt{3}-2$$

Steps for Completing the Square

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=\sqrt{3} x+2=-\sqrt{3}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{3}-2 x=-\sqrt{3}-2$$