# How do you solve #2x²+3x=5 # using the quadratic formula?

Bring the equation to standard form:

Use the improved quadratic formula (Socratic Search)(https://tutor.hix.ai):

There are 2 real roots:

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To solve the equation (2x^2 + 3x = 5) using the quadratic formula, you first need to rearrange the equation into the standard form (ax^2 + bx + c = 0), where (a), (b), and (c) are coefficients. In this case, (a = 2), (b = 3), and (c = -5). Then, apply the quadratic formula:

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

Substitute the values of (a), (b), and (c) into the formula:

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

Calculate the discriminant:

[b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49]

Plug the discriminant back into the quadratic formula:

[x = \frac{{-3 \pm \sqrt{{49}}}}{{4}}]

This gives two solutions:

[x_1 = \frac{{-3 + 7}}{{4}} = \frac{4}{4} = 1]

[x_2 = \frac{{-3 - 7}}{{4}} = \frac{-10}{4} = -\frac{5}{2}]

Therefore, the solutions to the equation are (x = 1) and (x = -\frac{5}{2}).

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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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