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

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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 
$$
x = -6 \quad \text{and} \quad x = \frac{2}{3}.
$$
The solutions to the equation are $$ x = -6 $$ and $$ x = \frac{2}{3} $$.

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

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