To factor the quadratic equation \(6x^2 + 5x - 6 = 0\), we start by looking for two numbers that multiply to \(ac = 6 \times -6 = -36\) and add to \(b = 5\).
The numbers that work are \(9\) and \(-4\) since:
- \(9 \times -4 = -36\)
- \(9 + (-4) = 5\)
We can now rewrite the middle term using these numbers: \[6x^2 + 9x - 4x - 6 = 0\]
Next, we group the terms: \[(6x^2 + 9x) + (-4x - 6) = 0\]
Now, factor by grouping: \[3x(2x + 3) - 2(2x + 3) = 0\]
Factoring out the common factor \((2x + 3)\): \[(2x + 3)(3x - 2) = 0\]
Now, we set each factor to zero:
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\(2x + 3 = 0\) \[2x = -3\] \[x = -\frac{3}{2}\]
-
\(3x - 2 = 0\) \[3x = 2\] \[x = \frac{2}{3}\]
The solutions to the quadratic equation \(6x^2 + 5x - 6 = 0\) are: \[x = -\frac{3}{2} \text{ and } x = \frac{2}{3}\]
From the provided options, the correct matching answers are:
- \(x = \frac{2}{3} \text{ (x is equal to 2 thirds)}\)
- \(x = -\frac{3}{2} \text{ (x is equal to negative 3 halves)}\) (not in the options, but mentioned for clarity)
Therefore, the final answer is:
- \(x = \frac{2}{3}\) (x is equal to 2 thirds)