Solve (m+3)^2=m+1

$$( m + 3 ) ^ { 2 } = m + 1$$

Solve for m

$$m=\frac{-5+\sqrt{7}i}{2}\approx -2.5+1.322875656i$$

$$m=\frac{-\sqrt{7}i-5}{2}\approx -2.5-1.322875656i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+3\right)^{2}.

$$m^{2}+6m+9=m+1$$

Subtract m from both sides.

$$m^{2}+6m+9-m=1$$

Combine 6m and -m to get 5m.

$$m^{2}+5m+9=1$$

Subtract 1 from both sides.

$$m^{2}+5m+9-1=0$$

Subtract 1 from 9 to get 8.

$$m^{2}+5m+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-5±\sqrt{5^{2}-4\times 8}}{2}$$

Square 5.

$$m=\frac{-5±\sqrt{25-4\times 8}}{2}$$

Multiply -4 times 8.

$$m=\frac{-5±\sqrt{25-32}}{2}$$

Add 25 to -32.

$$m=\frac{-5±\sqrt{-7}}{2}$$

Take the square root of -7.

$$m=\frac{-5±\sqrt{7}i}{2}$$

Now solve the equation m=\frac{-5±\sqrt{7}i}{2} when ± is plus. Add -5 to i\sqrt{7}.

$$m=\frac{-5+\sqrt{7}i}{2}$$

Now solve the equation m=\frac{-5±\sqrt{7}i}{2} when ± is minus. Subtract i\sqrt{7} from -5.

$$m=\frac{-\sqrt{7}i-5}{2}$$

The equation is now solved.

$$m=\frac{-5+\sqrt{7}i}{2} m=\frac{-\sqrt{7}i-5}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+3\right)^{2}.

$$m^{2}+6m+9=m+1$$

Subtract m from both sides.

$$m^{2}+6m+9-m=1$$

Combine 6m and -m to get 5m.

$$m^{2}+5m+9=1$$

Subtract 9 from both sides.

$$m^{2}+5m=1-9$$

Subtract 9 from 1 to get -8.

$$m^{2}+5m=-8$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}+5m+\left(\frac{5}{2}\right)^{2}=-8+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$m^{2}+5m+\frac{25}{4}=-8+\frac{25}{4}$$

Add -8 to \frac{25}{4}.

$$m^{2}+5m+\frac{25}{4}=-\frac{7}{4}$$

Factor m^{2}+5m+\frac{25}{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}.

$$\left(m+\frac{5}{2}\right)^{2}=-\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Simplify.

$$m+\frac{5}{2}=\frac{\sqrt{7}i}{2} m+\frac{5}{2}=-\frac{\sqrt{7}i}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$m=\frac{-5+\sqrt{7}i}{2} m=\frac{-\sqrt{7}i-5}{2}$$