Solve Using the Quadratic Formula n^2+3n-12=6

$n^{2} + 3 n - 12 = 6$

Step 1

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

Subtract $6$ from both sides of the equation.

$n^{2} + 3 n - 12 - 6 = 0$

Subtract $6$ from $- 12$ .

$n^{2} + 3 n - 18 = 0$

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 $a = 1$ , $b = 3$ , and $c = - 18$ into the quadratic formula and solve for $n$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 18)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$n = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 18}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 18$ .

Multiply $- 4$ by $1$ .

$n = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 18}}{2 \cdot 1}$

Multiply $- 4$ by $- 18$ .

$n = \frac{- 3 \pm \sqrt{9 + 72}}{2 \cdot 1}$

Add $9$ and $72$ .

$n = \frac{- 3 \pm \sqrt{81}}{2 \cdot 1}$

Rewrite $81$ as $9^{2}$ .

$n = \frac{- 3 \pm \sqrt{9^{2}}}{2 \cdot 1}$

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

$n = \frac{- 3 \pm 9}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{- 3 \pm 9}{2}$

Step 5

The final answer is the combination of both solutions.

$n = 3, - 6$