Solve (a+2)*a=-1

$$( a + 2 ) \cdot a = - 1$$

Solve for a

$$a=-1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply a+2 by a.

$$a^{2}+2a=-1$$

Add 1 to both sides.

$$a^{2}+2a+1=0$$

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

$$a=\frac{-2±\sqrt{2^{2}-4}}{2}$$

Square 2.

$$a=\frac{-2±\sqrt{4-4}}{2}$$

Add 4 to -4.

$$a=\frac{-2±\sqrt{0}}{2}$$

Take the square root of 0.

$$a=-\frac{2}{2}$$

Divide -2 by 2.

$$a=-1$$

Steps for Completing the Square

Use the distributive property to multiply a+2 by a.

$$a^{2}+2a=-1$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+2a+1^{2}=-1+1^{2}$$

Square 1.

$$a^{2}+2a+1=-1+1$$

Add -1 to 1.

$$a^{2}+2a+1=0$$

Factor a^{2}+2a+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$a+1=0 a+1=0$$

Subtract 1 from both sides of the equation.

$$a=-1 a=-1$$

The equation is now solved. Solutions are the same.

$$a=-1$$