How do you solve #2x^2+5x= - 4# using the quadratic formula?

Answer 1

(#-b + or -# #sqrt(#b^2 -4(a)(C)))÷(#2a) quadratic formula must equal zero

x

#b^2# ±√b−4ac


       2a

move 4 to the other side so

#2x^2 + 5x + 4=0#
#2x^2=a# #5x=b# #4=c#
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Answer 2

To solve the quadratic equation (2x^2 + 5x = -4) using the quadratic formula, first identify the coefficients: (a = 2), (b = 5), and (c = -4). Then, substitute these values into the quadratic formula:

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

Now, plug in the values:

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

Calculate inside the square root:

[5^2 - 4 \cdot 2 \cdot (-4) = 25 + 32 = 57]

Now, continue solving:

[x = \frac{{-5 \pm \sqrt{{57}}}}{{4}}]

Therefore, the solutions are:

[x = \frac{{-5 + \sqrt{{57}}}}{{4}} \text{ and } x = \frac{{-5 - \sqrt{{57}}}}{{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.

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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