Solve n+1=n^2-1

$$n + 1 = n ^ { 2 } - 1$$

Solve for n

$$n=-1$$

$$n=2$$

Steps Using Factoring By Grouping

Subtract n^{2} from both sides.

$$n+1-n^{2}=-1$$

Add 1 to both sides.

$$n+1-n^{2}+1=0$$

Add 1 and 1 to get 2.

$$n+2-n^{2}=0$$

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

$$-n^{2}+n+2=0$$

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

$$a+b=1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=2 b=-1$$

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

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

Factor out -n in the first and -1 in the second group.

$$-n\left(n-2\right)-\left(n-2\right)$$

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

$$\left(n-2\right)\left(-n-1\right)$$

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

$$n=2 n=-1$$

Steps Using the Quadratic Formula

Subtract n^{2} from both sides.

$$n+1-n^{2}=-1$$

Add 1 to both sides.

$$n+1-n^{2}+1=0$$

Add 1 and 1 to get 2.

$$n+2-n^{2}=0$$

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.

$$-n^{2}+n+2=0$$

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

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

Square 1.

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

Multiply -4 times -1.

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

Multiply 4 times 2.

$$n=\frac{-1±\sqrt{1+8}}{2\left(-1\right)}$$

Add 1 to 8.

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

Take the square root of 9.

$$n=\frac{-1±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{-1±3}{-2}$$

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

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

Divide 2 by -2.

$$n=-1$$

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

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

Divide -4 by -2.

$$n=2$$

The equation is now solved.

$$n=-1 n=2$$

Steps for Completing the Square

Subtract n^{2} from both sides.

$$n+1-n^{2}=-1$$

Subtract 1 from both sides.

$$n-n^{2}=-1-1$$

Subtract 1 from -1 to get -2.

$$n-n^{2}=-2$$

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.

$$-n^{2}+n=-2$$

Divide both sides by -1.

$$\frac{-n^{2}+n}{-1}=-\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\frac{1}{-1}n=-\frac{2}{-1}$$

Divide 1 by -1.

$$n^{2}-n=-\frac{2}{-1}$$

Divide -2 by -1.

$$n^{2}-n=2$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

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

Factor n^{2}-n+\frac{1}{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}.

$$\left(n-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$n=2 n=-1$$