How do you solve #6x^2+5x+1=0#?
The solutions are:
Factorising by splitting the middle term
Now we equate the factors with zero to obtain the solutions:
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To solve the quadratic equation (6x^2 + 5x + 1 = 0), you can use the quadratic formula. The quadratic formula states that for an equation in the form (ax^2 + bx + c = 0), the solutions for (x) are given by:
[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]
In your equation, (a = 6), (b = 5), and (c = 1). Substituting these values into the quadratic formula:
[x = \frac{{-5 \pm \sqrt{{5^2 - 4 \cdot 6 \cdot 1}}}}{{2 \cdot 6}}]
[x = \frac{{-5 \pm \sqrt{{25 - 24}}}}{{12}}]
[x = \frac{{-5 \pm \sqrt{{1}}}}{{12}}]
[x = \frac{{-5 \pm 1}}{{12}}]
This gives two possible solutions:
[x_1 = \frac{{-5 + 1}}{{12}} = \frac{{-4}}{{12}} = -\frac{{1}}{{3}}]
[x_2 = \frac{{-5 - 1}}{{12}} = \frac{{-6}}{{12}} = -\frac{{1}}{{2}}]
So, the solutions to the equation (6x^2 + 5x + 1 = 0) are (x = -\frac{{1}}{{3}}) and (x = -\frac{{1}}{{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.
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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