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How do you solve #5x^2-3x-3=0# using the quadratic formula?

Answer 1

Kennedy Adams

#x=(3+ sqrt(69))/10#

#x=(3- sqrt(69))/10#

The quadratic formula is #(-b± sqrt(b^2-4ac))/(2a)#

In the equation #5x^2-3x-3=0#

#a=5# #b=-3# #c=-3#

Consequently,

#(3± sqrt(-3^2-4xx5xx-3))/(2xx5)#

= #(3± sqrt(9--60))/(10)#

#x=(3± sqrt(69))/10#

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Answer 2

Avery Alexander

To solve the quadratic equation (5x^2 - 3x - 3 = 0) using the quadratic formula, (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 5), (b = -3), and (c = -3):

[x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 5 \cdot (-3)}}}}{{2 \cdot 5}}]

[x = \frac{{3 \pm \sqrt{{9 + 60}}}}{{10}}]

[x = \frac{{3 \pm \sqrt{{69}}}}{{10}}]

So the solutions are:

[x = \frac{{3 + \sqrt{{69}}}}{{10}} \text{ or } x = \frac{{3 - \sqrt{{69}}}}{{10}}]

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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.