What are the solutions to the equation? \( x=-6 \) and \( x=-\frac{2}{3} \) \( x=-6 \) and \( x=\frac{2}{3} \) \( x=6 \) and \( x=\frac{2}{3} \)

We start by checking the two candidate pairs of solutions. When we have a quadratic with two roots \( r_1 \) and \( r_2 \), we know that (up to a nonzero constant multiple) the equation can be written in factored form as \[ a(x - r_1)(x - r_2)=0. \] Two things to verify are the sum and product of the roots. If the two roots are \( r_1 \) and \( r_2 \), then by Vieta’s formulas for a quadratic \( ax^2+bx+c=0 \) (with \(a\ne 0\)) we have: \[ r_1 + r_2 = -\frac{b}{a} \quad \text{and} \quad r_1r_2 = \frac{c}{a}. \] Let’s compare the candidate answer options: 1. If the proposed solutions are \[ x = -6 \quad \text{and} \quad x = -\frac{2}{3}, \] then \[ r_1 + r_2 = -6 - \frac{2}{3} = -\frac{20}{3},\quad \text{and} \quad r_1r_2 = \left(-6\right)\left(-\frac{2}{3}\right) = 4. \] 2. If the proposed solutions are \[ x = -6 \quad \text{and} \quad x = \frac{2}{3}, \] then \[ r_1 + r_2 = -6 + \frac{2}{3} = -\frac{16}{3},\quad \text{and} \quad r_1r_2 = \left(-6\right)\left(\frac{2}{3}\right) = -4. \] 3. If the proposed solutions are \[ x = 6 \quad \text{and} \quad x = \frac{2}{3}, \] then \[ r_1 + r_2 = 6 + \frac{2}{3} = \frac{20}{3},\quad \text{and} \quad r_1r_2 = 6\left(\frac{2}{3}\right) = 4. \] Without the original equation explicitly given, one common scenario is that the quadratic’s coefficients are such that the factored form (after clearing fractions) is \[ 3(x + 6)\Bigl(x - \frac{2}{3}\Bigr) = 0. \] Expanding this gives: \[ 3\left[x^2 - \frac{2}{3}x + 6x - 4\right] = 3\left[x^2 + \frac{16}{3}x - 4 \right] = 3x^2 + 16x -12 = 0. \] Here the sum and product of the roots are: \[ r_1 + r_2 = -\frac{16}{3} \quad \text{and} \quad r_1r_2 = \frac{-12}{3} = -4, \] which exactly match the values from option 2. Thus, the solutions to the equation are \[ x = -6 \quad \text{and} \quad x = \frac{2}{3}. \]