Solve [-(m+2)/2]^2=2m+4

$$[ \frac { - ( m + 2 ) } { 2 } ] ^ { 2 } = 2 m + 4$$

Solve for m

$$m=-2$$

$$m=6$$

Steps Using the Quadratic Formula

To find the opposite of m+2, find the opposite of each term.

$$\left(\frac{-m-2}{2}\right)^{2}=2m+4$$

To raise \frac{-m-2}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-m-2\right)^{2}}{2^{2}}=2m+4$$

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

$$\frac{m^{2}+4m+4}{2^{2}}=2m+4$$

Calculate 2 to the power of 2 and get 4.

$$\frac{m^{2}+4m+4}{4}=2m+4$$

Divide each term of m^{2}+4m+4 by 4 to get \frac{1}{4}m^{2}+m+1.

$$\frac{1}{4}m^{2}+m+1=2m+4$$

Subtract 2m from both sides.

$$\frac{1}{4}m^{2}+m+1-2m=4$$

Combine m and -2m to get -m.

$$\frac{1}{4}m^{2}-m+1=4$$

Subtract 4 from both sides.

$$\frac{1}{4}m^{2}-m+1-4=0$$

Subtract 4 from 1 to get -3.

$$\frac{1}{4}m^{2}-m-3=0$$

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

$$m=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{4}\left(-3\right)}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$m=\frac{-\left(-1\right)±\sqrt{1-\left(-3\right)}}{2\times \frac{1}{4}}$$

Multiply -1 times -3.

$$m=\frac{-\left(-1\right)±\sqrt{1+3}}{2\times \frac{1}{4}}$$

Add 1 to 3.

$$m=\frac{-\left(-1\right)±\sqrt{4}}{2\times \frac{1}{4}}$$

Take the square root of 4.

$$m=\frac{-\left(-1\right)±2}{2\times \frac{1}{4}}$$

The opposite of -1 is 1.

$$m=\frac{1±2}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$m=\frac{1±2}{\frac{1}{2}}$$

Now solve the equation m=\frac{1±2}{\frac{1}{2}} when ± is plus. Add 1 to 2.

$$m=\frac{3}{\frac{1}{2}}$$

Divide 3 by \frac{1}{2} by multiplying 3 by the reciprocal of \frac{1}{2}.

$$m=6$$

Now solve the equation m=\frac{1±2}{\frac{1}{2}} when ± is minus. Subtract 2 from 1.

$$m=-\frac{1}{\frac{1}{2}}$$

Divide -1 by \frac{1}{2} by multiplying -1 by the reciprocal of \frac{1}{2}.

$$m=-2$$

The equation is now solved.

$$m=6 m=-2$$

Steps for Completing the Square

To find the opposite of m+2, find the opposite of each term.

$$\left(\frac{-m-2}{2}\right)^{2}=2m+4$$

To raise \frac{-m-2}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-m-2\right)^{2}}{2^{2}}=2m+4$$

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

$$\frac{m^{2}+4m+4}{2^{2}}=2m+4$$

Calculate 2 to the power of 2 and get 4.

$$\frac{m^{2}+4m+4}{4}=2m+4$$

Divide each term of m^{2}+4m+4 by 4 to get \frac{1}{4}m^{2}+m+1.

$$\frac{1}{4}m^{2}+m+1=2m+4$$

Subtract 2m from both sides.

$$\frac{1}{4}m^{2}+m+1-2m=4$$

Combine m and -2m to get -m.

$$\frac{1}{4}m^{2}-m+1=4$$

Subtract 1 from both sides.

$$\frac{1}{4}m^{2}-m=4-1$$

Subtract 1 from 4 to get 3.

$$\frac{1}{4}m^{2}-m=3$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}m^{2}-m}{\frac{1}{4}}=\frac{3}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$m^{2}+\left(-\frac{1}{\frac{1}{4}}\right)m=\frac{3}{\frac{1}{4}}$$

Divide -1 by \frac{1}{4} by multiplying -1 by the reciprocal of \frac{1}{4}.

$$m^{2}-4m=\frac{3}{\frac{1}{4}}$$

Divide 3 by \frac{1}{4} by multiplying 3 by the reciprocal of \frac{1}{4}.

$$m^{2}-4m=12$$

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.

$$m^{2}-4m+\left(-2\right)^{2}=12+\left(-2\right)^{2}$$

Square -2.

$$m^{2}-4m+4=12+4$$

Add 12 to 4.

$$m^{2}-4m+4=16$$

Factor m^{2}-4m+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}.

$$\left(m-2\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-2\right)^{2}}=\sqrt{16}$$

Simplify.

$$m-2=4 m-2=-4$$

Add 2 to both sides of the equation.

$$m=6 m=-2$$