Solve 4=(1-m)-(m-2)^2

$$4 = ( 1 - m ) - ( m - 2 ) ^ { 2 }$$

Solve for m

$$m=\frac{-\sqrt{19}i+3}{2}\approx 1.5-2.179449472i$$

$$m=\frac{3+\sqrt{19}i}{2}\approx 1.5+2.179449472i$$

Steps Using the Quadratic Formula

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

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

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

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

Combine -m and 4m to get 3m.

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

Subtract 4 from 1 to get -3.

$$4=-3+3m-m^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-3+3m-m^{2}=4$$

Subtract 4 from both sides.

$$-3+3m-m^{2}-4=0$$

Subtract 4 from -3 to get -7.

$$-7+3m-m^{2}=0$$

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.

$$-m^{2}+3m-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-7\right)}}{2\left(-1\right)}$$

Square 3.

$$m=\frac{-3±\sqrt{9-4\left(-1\right)\left(-7\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$m=\frac{-3±\sqrt{9+4\left(-7\right)}}{2\left(-1\right)}$$

Multiply 4 times -7.

$$m=\frac{-3±\sqrt{9-28}}{2\left(-1\right)}$$

Add 9 to -28.

$$m=\frac{-3±\sqrt{-19}}{2\left(-1\right)}$$

Take the square root of -19.

$$m=\frac{-3±\sqrt{19}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$m=\frac{-3±\sqrt{19}i}{-2}$$

Now solve the equation m=\frac{-3±\sqrt{19}i}{-2} when ± is plus. Add -3 to i\sqrt{19}.

$$m=\frac{-3+\sqrt{19}i}{-2}$$

Divide -3+i\sqrt{19} by -2.

$$m=\frac{-\sqrt{19}i+3}{2}$$

Now solve the equation m=\frac{-3±\sqrt{19}i}{-2} when ± is minus. Subtract i\sqrt{19} from -3.

$$m=\frac{-\sqrt{19}i-3}{-2}$$

Divide -3-i\sqrt{19} by -2.

$$m=\frac{3+\sqrt{19}i}{2}$$

The equation is now solved.

$$m=\frac{-\sqrt{19}i+3}{2} m=\frac{3+\sqrt{19}i}{2}$$

Steps for Completing the Square

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

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

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

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

Combine -m and 4m to get 3m.

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

Subtract 4 from 1 to get -3.

$$4=-3+3m-m^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-3+3m-m^{2}=4$$

Add 3 to both sides.

$$3m-m^{2}=4+3$$

Add 4 and 3 to get 7.

$$3m-m^{2}=7$$

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.

$$-m^{2}+3m=7$$

Divide both sides by -1.

$$\frac{-m^{2}+3m}{-1}=\frac{7}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$m^{2}+\frac{3}{-1}m=\frac{7}{-1}$$

Divide 3 by -1.

$$m^{2}-3m=\frac{7}{-1}$$

Divide 7 by -1.

$$m^{2}-3m=-7$$

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.

$$m^{2}-3m+\left(-\frac{3}{2}\right)^{2}=-7+\left(-\frac{3}{2}\right)^{2}$$

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

$$m^{2}-3m+\frac{9}{4}=-7+\frac{9}{4}$$

Add -7 to \frac{9}{4}.

$$m^{2}-3m+\frac{9}{4}=-\frac{19}{4}$$

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}.

$$\left(m-\frac{3}{2}\right)^{2}=-\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{19}{4}}$$

Simplify.

$$m-\frac{3}{2}=\frac{\sqrt{19}i}{2} m-\frac{3}{2}=-\frac{\sqrt{19}i}{2}$$

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

$$m=\frac{3+\sqrt{19}i}{2} m=\frac{-\sqrt{19}i+3}{2}$$