Solve Using the Quadratic Formula 3n^2-2n-5=0

$3 n^{2} - 2 n - 5 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 3$ , $b = - 2$ , and $c = - 5$ into the quadratic formula and solve for $n$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (3 \cdot - 5)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

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

$n = \frac{2 \pm \sqrt{4 - 4 \cdot 3 \cdot - 5}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 5$ .

Multiply $- 4$ by $3$ .

$n = \frac{2 \pm \sqrt{4 - 12 \cdot - 5}}{2 \cdot 3}$

Multiply $- 12$ by $- 5$ .

$n = \frac{2 \pm \sqrt{4 + 60}}{2 \cdot 3}$

Add $4$ and $60$ .

$n = \frac{2 \pm \sqrt{64}}{2 \cdot 3}$

Rewrite $64$ as $8^{2}$ .

$n = \frac{2 \pm \sqrt{8^{2}}}{2 \cdot 3}$

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

$n = \frac{2 \pm 8}{2 \cdot 3}$

Multiply $2$ by $3$ .

$n = \frac{2 \pm 8}{6}$

Simplify $\frac{2 \pm 8}{6}$ .

$n = \frac{1 \pm 4}{3}$

Step 4

The final answer is the combination of both solutions.

$n = \frac{5}{3}, - 1$