Solve -8m-(m-2)^2/4=0

$$\frac { - 8 m - ( m - 2 ) ^ { 2 } } { 4 } = 0$$

Solve for m

$$m=-2$$

Steps Using Factoring By Grouping

Multiply both sides by 4. Anything times zero gives zero.

$$-8m-\left(m-2\right)^{2}=0$$

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

$$-8m-\left(m^{2}-4m+4\right)=0$$

To find the opposite of m^{2}-4m+4, find the opposite of each term.

$$-8m-m^{2}+4m-4=0$$

Combine -8m and 4m to get -4m.

$$-4m-m^{2}-4=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-m^{2}-4m-4=0$$

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

$$a+b=-4 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 -m^{2}-4m-4 as \left(-m^{2}-2m\right)+\left(-2m-4\right).

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

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

$$m\left(-m-2\right)+2\left(-m-2\right)$$

Factor out common term -m-2 by using distributive property.

$$\left(-m-2\right)\left(m+2\right)$$

To find equation solutions, solve -m-2=0 and m+2=0.

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

Steps Using the Quadratic Formula

Multiply both sides by 4. Anything times zero gives zero.

$$-8m-\left(m-2\right)^{2}=0$$

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

$$-8m-\left(m^{2}-4m+4\right)=0$$

To find the opposite of m^{2}-4m+4, find the opposite of each term.

$$-8m-m^{2}+4m-4=0$$

Combine -8m and 4m to get -4m.

$$-4m-m^{2}-4=0$$

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.

$$-m^{2}-4m-4=0$$

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}.

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

Square -4.

$$m=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$m=\frac{-\left(-4\right)±\sqrt{16+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$m=\frac{-\left(-4\right)±\sqrt{16-16}}{2\left(-1\right)}$$

Add 16 to -16.

$$m=\frac{-\left(-4\right)±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

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

The opposite of -4 is 4.

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

Multiply 2 times -1.

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

Divide 4 by -2.

$$m=-2$$

Steps for Completing the Square

Multiply both sides by 4. Anything times zero gives zero.

$$-8m-\left(m-2\right)^{2}=0$$

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

$$-8m-\left(m^{2}-4m+4\right)=0$$

To find the opposite of m^{2}-4m+4, find the opposite of each term.

$$-8m-m^{2}+4m-4=0$$

Combine -8m and 4m to get -4m.

$$-4m-m^{2}-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

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

Divide 4 by -1.

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

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.

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

Square 2.

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

Add -4 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$m+2=0 m+2=0$$

Subtract 2 from both sides of the equation.

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

The equation is now solved. Solutions are the same.

$$m=-2$$