# How do you solve #2x^2+3x-2=0#?

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To solve the quadratic equation (2x^2 + 3x - 2 = 0), you can use the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where (a), (b), and (c) are the coefficients of the quadratic equation in the form (ax^2 + bx + c = 0).

For the given equation (2x^2 + 3x - 2 = 0), the coefficients are (a = 2), (b = 3), and (c = -2). Plugging these values into the quadratic formula, we get:

[x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}]

Simplifying inside the square root gives:

[x = \frac{-3 \pm \sqrt{9 + 16}}{4}] [x = \frac{-3 \pm \sqrt{25}}{4}] [x = \frac{-3 \pm 5}{4}]

So, the solutions are:

[x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}]

[x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2]

Therefore, the solutions to the equation (2x^2 + 3x - 2 = 0) are (x = \frac{1}{2}) and (x = -2).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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