Solve -left(1-mright)^2+m+2=3

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

Solve for m

$$m=1$$

$$m=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-m\right)^{2}.

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

To find the opposite of 1-2m+m^{2}, find the opposite of each term.

$$-1+2m-m^{2}+m+2=3$$

Combine 2m and m to get 3m.

$$-1+3m-m^{2}+2=3$$

Add -1 and 2 to get 1.

$$1+3m-m^{2}=3$$

Subtract 3 from both sides.

$$1+3m-m^{2}-3=0$$

Subtract 3 from 1 to get -2.

$$-2+3m-m^{2}=0$$

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

$$-m^{2}+3m-2=0$$

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

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

Rewrite -m^{2}+3m-2 as \left(-m^{2}+2m\right)+\left(m-2\right).

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

Factor out -m in -m^{2}+2m.

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

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

$$\left(m-2\right)\left(-m+1\right)$$

To find equation solutions, solve m-2=0 and -m+1=0.

$$m=2 m=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-m\right)^{2}.

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

To find the opposite of 1-2m+m^{2}, find the opposite of each term.

$$-1+2m-m^{2}+m+2=3$$

Combine 2m and m to get 3m.

$$-1+3m-m^{2}+2=3$$

Add -1 and 2 to get 1.

$$1+3m-m^{2}=3$$

Subtract 3 from both sides.

$$1+3m-m^{2}-3=0$$

Subtract 3 from 1 to get -2.

$$-2+3m-m^{2}=0$$

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.

$$-m^{2}+3m-2=0$$

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

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

Square 3.

$$m=\frac{-3±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$m=\frac{-3±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$m=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

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

Take the square root of 1.

$$m=\frac{-3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$m=\frac{-3±1}{-2}$$

Now solve the equation m=\frac{-3±1}{-2} when ± is plus. Add -3 to 1.

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

Divide -2 by -2.

$$m=1$$

Now solve the equation m=\frac{-3±1}{-2} when ± is minus. Subtract 1 from -3.

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

Divide -4 by -2.

$$m=2$$

The equation is now solved.

$$m=1 m=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-m\right)^{2}.

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

To find the opposite of 1-2m+m^{2}, find the opposite of each term.

$$-1+2m-m^{2}+m+2=3$$

Combine 2m and m to get 3m.

$$-1+3m-m^{2}+2=3$$

Add -1 and 2 to get 1.

$$1+3m-m^{2}=3$$

Subtract 1 from both sides.

$$3m-m^{2}=3-1$$

Subtract 1 from 3 to get 2.

$$3m-m^{2}=2$$

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.

$$-m^{2}+3m=2$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

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

Divide 2 by -1.

$$m^{2}-3m=-2$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-3m+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$m^{2}-3m+\frac{9}{4}=\frac{1}{4}$$

Factor m^{2}-3m+\frac{9}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$m=2 m=1$$