Solve -0.5(n+2)^2=0

$$- 0.5 ( n + 2 ) ^ { 2 } = 0$$

Solve for n

$$n=-2$$

Steps Using Factoring

Divide both sides by -0.5. Zero divided by any non-zero number gives zero.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+2\right)^{2}.

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

To solve the equation, factor n^{2}+4n+4 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.

$$\left(n+2\right)\left(n+2\right)$$

Rewrite as a binomial square.

$$\left(n+2\right)^{2}$$

To find equation solution, solve n+2=0.

$$n=-2$$

Steps Using Factoring By Grouping

Divide both sides by -0.5. Zero divided by any non-zero number gives zero.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+2\right)^{2}.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn+4. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite n^{2}+4n+4 as \left(n^{2}+2n\right)+\left(2n+4\right).

$$\left(n^{2}+2n\right)+\left(2n+4\right)$$

Factor out n in the first and 2 in the second group.

$$n\left(n+2\right)+2\left(n+2\right)$$

Factor out common term n+2 by using distributive property.

$$\left(n+2\right)\left(n+2\right)$$

Rewrite as a binomial square.

$$\left(n+2\right)^{2}$$

To find equation solution, solve n+2=0.

$$n=-2$$

Steps Using the Quadratic Formula

Divide both sides by -0.5. Zero divided by any non-zero number gives zero.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(n+2\right)^{2}.

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

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}.

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

Square 4.

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

Multiply -4 times 4.

$$n=\frac{-4±\sqrt{16-16}}{2}$$

Add 16 to -16.

$$n=\frac{-4±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide -4 by 2.

$$n=-2$$

Steps for Completing the Square

Divide both sides by -0.5. Zero divided by any non-zero number gives zero.

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

Take the square root of both sides of the equation.

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

Simplify.

$$n+2=0 n+2=0$$

Subtract 2 from both sides of the equation.

$$n=-2 n=-2$$

The equation is now solved. Solutions are the same.

$$n=-2$$