Solve Using the Quadratic Formula n^2-5=-4

$n^{2} - 5 = - 4$

Step 1

Move all terms to the left side of the equation and simplify.

Add $4$ to both sides of the equation.

$n^{2} - 5 + 4 = 0$

Add $- 5$ and $4$ .

$n^{2} - 1 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 0$ , and $c = - 1$ into the quadratic formula and solve for $n$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$n = \frac{0 \pm \sqrt{0 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

$n = \frac{0 \pm \sqrt{0 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$n = \frac{0 \pm \sqrt{0 + 4}}{2 \cdot 1}$

Add $0$ and $4$ .

$n = \frac{0 \pm \sqrt{4}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$n = \frac{0 \pm \sqrt{2^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$n = \frac{0 \pm 2}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{0 \pm 2}{2}$

Simplify $\frac{0 \pm 2}{2}$ .

$n = \pm 1$

Step 5

The final answer is the combination of both solutions.

$n = 1, - 1$