138. \( \frac{1}{3} n^{2}+n=-\frac{1}{2} \)

To solve the equation \( \frac{1}{3} n^{2} + n = -\frac{1}{2} \), first, let's eliminate the fraction by multiplying every term by 6 to simplify calculations: \[ 6 \left(\frac{1}{3} n^{2} + n\right) = 6 \left(-\frac{1}{2}\right) \] This results in: \[ 2n^{2} + 6n = -3 \] Now, rearranging it gives: \[ 2n^{2} + 6n + 3 = 0 \] Next, using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 2, b = 6, c = 3 \): 1. Calculate the discriminant: \( b^2 - 4ac = 6^2 - 4 \cdot 2 \cdot 3 = 36 - 24 = 12 \) 2. Now plug into the formula: \( n = \frac{-6 \pm \sqrt{12}}{2 \cdot 2} \) \( n = \frac{-6 \pm 2\sqrt{3}}{4} \) \( n = \frac{-3 \pm \sqrt{3}}{2} \) Thus, the solutions are: \[ n = \frac{-3 + \sqrt{3}}{2} \quad \text{and} \quad n = \frac{-3 - \sqrt{3}}{2} \] These two solutions represent the values of \( n \) that satisfy the original equation!