Solve x^2+4x+4=7

$$x ^ { 2 } + 4 x + 4 = 7$$

Solve for x (complex solution)

$$x=\sqrt{7}-2\approx 0.645751311$$

$$x=-\left(\sqrt{7}+2\right)\approx -4.645751311$$

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.

$$x^{2}+4x+4=7$$

Subtract 7 from both sides of the equation.

$$x^{2}+4x+4-7=7-7$$

Subtracting 7 from itself leaves 0.

$$x^{2}+4x+4-7=0$$

Subtract 7 from 4.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x=\frac{-4±\sqrt{16+12}}{2}$$

Add 16 to 12.

$$x=\frac{-4±\sqrt{28}}{2}$$

Take the square root of 28.

$$x=\frac{-4±2\sqrt{7}}{2}$$

Now solve the equation x=\frac{-4±2\sqrt{7}}{2} when ± is plus. Add -4 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-4}{2}$$

Divide -4+2\sqrt{7} by 2.

$$x=\sqrt{7}-2$$

Now solve the equation x=\frac{-4±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from -4.

$$x=\frac{-2\sqrt{7}-4}{2}$$

Divide -4-2\sqrt{7} by 2.

$$x=-\sqrt{7}-2$$

The equation is now solved.

$$x=\sqrt{7}-2 x=-\sqrt{7}-2$$

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}.

$$\left(x+2\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{7}$$

Simplify.

$$x+2=\sqrt{7} x+2=-\sqrt{7}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{7}-2 x=-\sqrt{7}-2$$

Solve for x

$$x=\sqrt{7}-2\approx 0.645751311$$

$$x=-\sqrt{7}-2\approx -4.645751311$$

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.

$$x^{2}+4x+4=7$$

Subtract 7 from both sides of the equation.

$$x^{2}+4x+4-7=7-7$$

Subtracting 7 from itself leaves 0.

$$x^{2}+4x+4-7=0$$

Subtract 7 from 4.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x=\frac{-4±\sqrt{16+12}}{2}$$

Add 16 to 12.

$$x=\frac{-4±\sqrt{28}}{2}$$

Take the square root of 28.

$$x=\frac{-4±2\sqrt{7}}{2}$$

Now solve the equation x=\frac{-4±2\sqrt{7}}{2} when ± is plus. Add -4 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-4}{2}$$

Divide -4+2\sqrt{7} by 2.

$$x=\sqrt{7}-2$$

Now solve the equation x=\frac{-4±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from -4.

$$x=\frac{-2\sqrt{7}-4}{2}$$

Divide -4-2\sqrt{7} by 2.

$$x=-\sqrt{7}-2$$

The equation is now solved.

$$x=\sqrt{7}-2 x=-\sqrt{7}-2$$

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}.

$$\left(x+2\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{7}$$

Simplify.

$$x+2=\sqrt{7} x+2=-\sqrt{7}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{7}-2 x=-\sqrt{7}-2$$