Solve x(x+4)+5=1

$$x ( x + 4 ) + 5 = 1$$

Solve for x

$$x=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x+4.

$$x^{2}+4x+5=1$$

Subtract 1 from both sides.

$$x^{2}+4x+5-1=0$$

Subtract 1 from 5 to get 4.

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

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

$$x=\frac{-4±\sqrt{4^{2}-4\times 4}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16}}{2}$$

Add 16 to -16.

$$x=\frac{-4±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide -4 by 2.

$$x=-2$$

Steps for Completing the Square

Use the distributive property to multiply x by x+4.

$$x^{2}+4x+5=1$$

Subtract 5 from both sides.

$$x^{2}+4x=1-5$$

Subtract 5 from 1 to get -4.

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

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.

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

Square 2.

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

Add -4 to 4.

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

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}.

$$\left(x+2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+2=0 x+2=0$$

Subtract 2 from both sides of the equation.

$$x=-2 x=-2$$

The equation is now solved. Solutions are the same.

$$x=-2$$