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Answer:
To solve the equation \((3x-8)(x+2) = 2x^2\), we can start by expanding the left-hand side:
\[
(3x - 8)(x + 2) = 3x(x + 2) - 8(x + 2)
\]
\[
= 3x^2 + 6x - 8x - 16
\]
\[
= 3x^2 - 2x - 16
\]
Now, we set the expanded form equal to the right-hand side of the original equation:
\[
3x^2 - 2x - 16 = 2x^2
\]
Next, we subtract \(2x^2\) from both sides to set the equation to zero:
\[
3x^2 - 2x - 16 - 2x^2 = 0
\]
\[
x^2 - 2x - 16 = 0
\]
Now we have a quadratic equation. We can solve for \(x\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -16\):
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-16)}}{2(1)}
\]
\[
x = \frac{2 \pm \sqrt{4 + 64}}{2}
\]
\[
x = \frac{2 \pm \sqrt{68}}{2}
\]
\[
x = \frac{2 \pm 2\sqrt{17}}{2}
\]
\[
x = 1 \pm \sqrt{17}
\]
So, the solutions to the equation are:
\[
x = 1 + \sqrt{17} \quad \text{and} \quad x = 1 - \sqrt{17}
\]