Solve (x+3)^2-(x+3)=1

$$( x + 3 ) ^ { 2 } - ( x + 3 ) = 1$$

Solve for x

$$x=\frac{\sqrt{5}-5}{2}\approx -1.381966011$$

$$x=\frac{-\sqrt{5}-5}{2}\approx -3.618033989$$

Steps Using the Quadratic Formula

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

$$x^{2}+6x+9-\left(x+3\right)=1$$

To find the opposite of x+3, find the opposite of each term.

$$x^{2}+6x+9-x-3=1$$

Combine 6x and -x to get 5x.

$$x^{2}+5x+9-3=1$$

Subtract 3 from 9 to get 6.

$$x^{2}+5x+6=1$$

Subtract 1 from both sides.

$$x^{2}+5x+6-1=0$$

Subtract 1 from 6 to get 5.

$$x^{2}+5x+5=0$$

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

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

Square 5.

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

Multiply -4 times 5.

$$x=\frac{-5±\sqrt{25-20}}{2}$$

Add 25 to -20.

$$x=\frac{-5±\sqrt{5}}{2}$$

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

$$x=\frac{\sqrt{5}-5}{2}$$

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

$$x=\frac{-\sqrt{5}-5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}-5}{2} x=\frac{-\sqrt{5}-5}{2}$$

Steps for Completing the Square

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

$$x^{2}+6x+9-\left(x+3\right)=1$$

To find the opposite of x+3, find the opposite of each term.

$$x^{2}+6x+9-x-3=1$$

Combine 6x and -x to get 5x.

$$x^{2}+5x+9-3=1$$

Subtract 3 from 9 to get 6.

$$x^{2}+5x+6=1$$

Subtract 6 from both sides.

$$x^{2}+5x=1-6$$

Subtract 6 from 1 to get -5.

$$x^{2}+5x=-5$$

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.

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

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

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

Add -5 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{5}{4}$$

Factor x^{2}+5x+\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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{5}}{2} x+\frac{5}{2}=-\frac{\sqrt{5}}{2}$$

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

$$x=\frac{\sqrt{5}-5}{2} x=\frac{-\sqrt{5}-5}{2}$$