Solve Using the Quadratic Formula (6n+1)^2+3(6n+1)-4=0

${(6 n + 1)}^{2} + 3 (6 n + 1) - 4 = 0$

Step 1

Simplify the left side.

Simplify ${(6 n + 1)}^{2} + 3 (6 n + 1) - 4$ .

Simplify each term.

Apply the distributive property.

${(6 n + 1)}^{2} + 3 (6 n) + 3 \cdot 1 - 4 = 0$

Multiply $6$ by $3$ .

${(6 n + 1)}^{2} + 18 n + 3 \cdot 1 - 4 = 0$

Multiply $3$ by $1$ .

${(6 n + 1)}^{2} + 18 n + 3 - 4 = 0$

Subtract $4$ from $3$ .

${(6 n + 1)}^{2} + 18 n - 1 = 0$

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 $a = 36$ , $b = 30$ , and $c = 0$ into the quadratic formula and solve for $n$ .

$\frac{- 30 \pm \sqrt{30^{2} - 4 \cdot (36 \cdot 0)}}{2 \cdot 36}$

Step 4

Simplify.

Simplify the numerator.

Raise $30$ to the power of $2$ .

$n = \frac{- 30 \pm \sqrt{900 - 4 \cdot 36 \cdot 0}}{2 \cdot 36}$

Multiply $- 4 \cdot 36 \cdot 0$ .

Multiply $- 4$ by $36$ .

$n = \frac{- 30 \pm \sqrt{900 - 144 \cdot 0}}{2 \cdot 36}$

Multiply $- 144$ by $0$ .

$n = \frac{- 30 \pm \sqrt{900 + 0}}{2 \cdot 36}$

Add $900$ and $0$ .

$n = \frac{- 30 \pm \sqrt{900}}{2 \cdot 36}$

Rewrite $900$ as $30^{2}$ .

$n = \frac{- 30 \pm \sqrt{30^{2}}}{2 \cdot 36}$

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

$n = \frac{- 30 \pm 30}{2 \cdot 36}$

Multiply $2$ by $36$ .

$n = \frac{- 30 \pm 30}{72}$

Simplify $\frac{- 30 \pm 30}{72}$ .

$n = \frac{- 5 \pm 5}{12}$

Step 5

The final answer is the combination of both solutions.

$n = 0, - \frac{5}{6}$