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

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

Step 1

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

Subtract $2$ from both sides of the equation.

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

Subtract $2$ from $- 4$ .

$2 n^{2} - n - 6 = 0$

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 $a = 2$ , $b = - 1$ , and $c = - 6$ into the quadratic formula and solve for $n$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (2 \cdot - 6)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$n = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot - 6}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 6$ .

Multiply $- 4$ by $2$ .

$n = \frac{1 \pm \sqrt{1 - 8 \cdot - 6}}{2 \cdot 2}$

Multiply $- 8$ by $- 6$ .

$n = \frac{1 \pm \sqrt{1 + 48}}{2 \cdot 2}$

Add $1$ and $48$ .

$n = \frac{1 \pm \sqrt{49}}{2 \cdot 2}$

Rewrite $49$ as $7^{2}$ .

$n = \frac{1 \pm \sqrt{7^{2}}}{2 \cdot 2}$

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

$n = \frac{1 \pm 7}{2 \cdot 2}$

Multiply $2$ by $2$ .

$n = \frac{1 \pm 7}{4}$

Step 5

The final answer is the combination of both solutions.

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