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How do you solve #2x ^ { 2} + 10x - 168= 0#?

Answer 1

Hazel Anglin

#**x = -12, 7**#

#2x^2+10x-168=0#. Dividing by 2, #x^2+5x-84=0#

#x^2+12x-7x-84=0#

#x(x+12)-7(x+12)=0#

#(x+12)(x-7)=0# #x = -12, 7#

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Answer 2

Abigail Allen

To solve (2x^2 + 10x - 168 = 0), you can use the quadratic formula:

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

In this equation, (a = 2), (b = 10), and (c = -168). Substituting these values into the quadratic formula, you can calculate the roots of the equation.

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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.