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How do you solve equation with quadratic formula #3n^2-4n-15=0#?

Answer 1

Eva Adamson

#n_1=-5/3; n_2=3#

Since b-4 is even, you can use the formula:

#x=(-b/2+-sqrt((b/2)^2-ac))/a#, so you have

#n=(-(-4/2)+-sqrt((4/2)^2-3*(-15)))/3# #=(2+-sqrt(4+45))/3# #=(2+-7)/3# #n_1=-5/3; n_2=3#

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Answer 2

Julia Adams

To solve the equation (3n^2 - 4n - 15 = 0) using the quadratic formula, first identify the coefficients:

(a = 3) (b = -4) (c = -15)

Then, substitute these values into the quadratic formula:

[n = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]

[n = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(3)(-15)}}}}{{2(3)}}]

[n = \frac{{4 \pm \sqrt{{16 + 180}}}}{{6}}]

[n = \frac{{4 \pm \sqrt{{196}}}}{{6}}]

[n = \frac{{4 \pm 14}}{{6}}]

This yields two possible solutions:

[n_1 = \frac{{4 + 14}}{{6}} = \frac{{18}}{{6}} = 3]

[n_2 = \frac{{4 - 14}}{{6}} = \frac{{-10}}{{6}} = -\frac{{5}}{{3}}]

So, the solutions to the equation are (n = 3) and (n = -\frac{{5}}{{3}}).

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Answer from HIX Tutor

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.