Solve Using the Quadratic Formula (3n-2)(4n+1)=0

$(3 n - 2) (4 n + 1) = 0$

Step 1

Simplify the left side.

Simplify $(3 n - 2) (4 n + 1)$ .

Expand $(3 n - 2) (4 n + 1)$ using the FOIL Method.

Apply the distributive property.

$3 n (4 n + 1) - 2 (4 n + 1) = 0$

Apply the distributive property.

$3 n (4 n) + 3 n \cdot 1 - 2 (4 n + 1) = 0$

Apply the distributive property.

$3 n (4 n) + 3 n \cdot 1 - 2 (4 n) - 2 \cdot 1 = 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 \cdot 4 n \cdot n + 3 n \cdot 1 - 2 (4 n) - 2 \cdot 1 = 0$

Multiply $n$ by $n$ by adding the exponents.

Move $n$ .

$3 \cdot 4 (n \cdot n) + 3 n \cdot 1 - 2 (4 n) - 2 \cdot 1 = 0$

Multiply $n$ by $n$ .

$3 \cdot 4 n^{2} + 3 n \cdot 1 - 2 (4 n) - 2 \cdot 1 = 0$

Multiply $3$ by $4$ .

$12 n^{2} + 3 n \cdot 1 - 2 (4 n) - 2 \cdot 1 = 0$

Multiply $3$ by $1$ .

$12 n^{2} + 3 n - 2 (4 n) - 2 \cdot 1 = 0$

Multiply $4$ by $- 2$ .

$12 n^{2} + 3 n - 8 n - 2 \cdot 1 = 0$

Multiply $- 2$ by $1$ .

$12 n^{2} + 3 n - 8 n - 2 = 0$

Subtract $8 n$ from $3 n$ .

$12 n^{2} - 5 n - 2 = 0$

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 $a = 12$ , $b = - 5$ , and $c = - 2$ into the quadratic formula and solve for $n$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (12 \cdot - 2)}}{2 \cdot 12}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$n = \frac{5 \pm \sqrt{25 - 4 \cdot 12 \cdot - 2}}{2 \cdot 12}$

Multiply $- 4 \cdot 12 \cdot - 2$ .

Multiply $- 4$ by $12$ .

$n = \frac{5 \pm \sqrt{25 - 48 \cdot - 2}}{2 \cdot 12}$

Multiply $- 48$ by $- 2$ .

$n = \frac{5 \pm \sqrt{25 + 96}}{2 \cdot 12}$

Add $25$ and $96$ .

$n = \frac{5 \pm \sqrt{121}}{2 \cdot 12}$

Rewrite $121$ as $11^{2}$ .

$n = \frac{5 \pm \sqrt{11^{2}}}{2 \cdot 12}$

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

$n = \frac{5 \pm 11}{2 \cdot 12}$

Multiply $2$ by $12$ .

$n = \frac{5 \pm 11}{24}$

Step 5

The final answer is the combination of both solutions.

$n = \frac{2}{3}, - \frac{1}{4}$

Solve Using the Quadratic Formula (3n-2)(4n+1)=0

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