# How do you solve using the quadratic formula #3x(x - 2) = 1#?

f(x) = 3x^2 - 6x - 1 = 0

#D = d^2 = b^2 - 4ac = 36 + 12 = 48 = 16.3 -> d = +-4.sqr3#

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To solve the equation (3x(x - 2) = 1) using the quadratic formula, first, we need to rewrite the equation in standard quadratic form (ax^2 + bx + c = 0). Then, we can identify (a), (b), and (c) and substitute them into the quadratic formula:

(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}})

Given equation: (3x(x - 2) = 1)

Expanding and rearranging the equation:

(3x^2 - 6x - 1 = 0)

Now, we can identify (a), (b), and (c):

(a = 3), (b = -6), (c = -1)

Substituting into the quadratic formula:

(x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 3 \cdot (-1)}}}}{{2 \cdot 3}})

(x = \frac{{6 \pm \sqrt{{36 + 12}}}}{{6}})

(x = \frac{{6 \pm \sqrt{{48}}}}{{6}})

(x = \frac{{6 \pm 4\sqrt{3}}}{{6}})

Now, we can simplify further:

(x = \frac{{3 \pm 2\sqrt{3}}}{{3}})

Therefore, the solutions to the equation (3x(x - 2) = 1) using the quadratic formula are:

(x = \frac{{3 + 2\sqrt{3}}}{{3}}) and (x = \frac{{3 - 2\sqrt{3}}}{{3}})

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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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