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How do you solve #2x^2 - 7x - 3 = 0 #?

Answer 1

#x=(7+-sqrt(73))/4#

An examination of the factors of #2# and #-3# reveals that there are no integer solutions.

Therefore, we will use the quadratic root formula #color(white)("XXX")x= (-b+-sqrt(b^2-4ac))/(2a)# #color(white)("XXXXXXXXXXX")#for a quadratic in the general form #ax^2+bx+c=0#

In this case #color(white)("XXX")x = (7+-sqrt(7^2-4(2)(-3)))/(2(2))#

#color(white)("XXXX")=(7+-sqrt(49+24))/4#

#color(white)("XXXX")=(7+-sqrt(73))/4#

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Answer 2

Savannah Anderson

To solve the quadratic equation (2x^2 - 7x - 3 = 0), you can use the quadratic formula, which is (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 2), (b = -7), and (c = -3):

[x = \frac{{-(-7) \pm \sqrt{{(-7)^2 - 4(2)(-3)}}}}{{2(2)}}]

[x = \frac{{7 \pm \sqrt{{49 + 24}}}}{4}]

[x = \frac{{7 \pm \sqrt{73}}}{4}]

So, the solutions are (x = \frac{7 + \sqrt{73}}{4}) and (x = \frac{7 - \sqrt{73}}{4}).

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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.