How do you solve #2x^2+5x=0# using the quadratic formula?
Express in the form
#x = 0" "# or#" "x = 5/2#
Given:
We can write this equation as follows:
This takes the following form:
Its roots can be found using the quadratic formula:
Thus, one root is:
and the other is:
In this case, you really wouldn't want to use the quadratic formula.
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To solve the quadratic equation (2x^2 + 5x = 0) using the quadratic formula, which is (x = \frac{{b \pm \sqrt{{b^2  4ac}}}}{{2a}}), where (a), (b), and (c) are the coefficients of the quadratic equation (ax^2 + bx + c = 0), follow these steps:

Identify the values of (a), (b), and (c): (a = 2), (b = 5), and (c = 0).

Substitute these values into the quadratic formula: (x = \frac{{5 \pm \sqrt{{5^2  4 \cdot 2 \cdot 0}}}}{{2 \cdot 2}}).

Simplify inside the square root: (x = \frac{{5 \pm \sqrt{{25}}}}{{4}}).

Take the square root of (25): (x = \frac{{5 \pm 5}}{{4}}).

Solve for both possible values of (x): (x_1 = \frac{{5 + 5}}{{4}} = \frac{0}{4} = 0), (x_2 = \frac{{5  5}}{{4}} = \frac{{10}}{{4}} = \frac{5}{2}).
Therefore, the solutions to the equation (2x^2 + 5x = 0) are (x = 0) and (x = \frac{5}{2}).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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