# How do you solve #x^2+3x-10=0#?

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To solve the quadratic equation (x^2 + 3x - 10 = 0), you can use 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.

For this equation, (a = 1), (b = 3), and (c = -10).

Plugging these values into the quadratic formula:

[x = \frac{{-3 \pm \sqrt{{3^2 - 4(1)(-10)}}}}{{2(1)}}]

[x = \frac{{-3 \pm \sqrt{{9 + 40}}}}{2}]

[x = \frac{{-3 \pm \sqrt{{49}}}}{2}]

[x = \frac{{-3 \pm 7}}{2}]

This gives two possible solutions:

[x_1 = \frac{{-3 + 7}}{2} = 2]

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

So, the solutions to the equation (x^2 + 3x - 10 = 0) are (x = 2) and (x = -5).

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To solve the quadratic equation (x^2 + 3x - 10 = 0), you can use the quadratic formula:

[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).

For the given equation (x^2 + 3x - 10 = 0): [a = 1] [b = 3] [c = -10]

Substitute these values into the quadratic formula and solve for (x):

[x = \frac{{-3 \pm \sqrt{{(3)^2 - 4(1)(-10)}}}}{{2(1)}}] [x = \frac{{-3 \pm \sqrt{{9 + 40}}}}{{2}}] [x = \frac{{-3 \pm \sqrt{{49}}}}{{2}}] [x = \frac{{-3 \pm 7}}{{2}}]

This gives us two solutions: [x_1 = \frac{{-3 + 7}}{{2}} = 2] [x_2 = \frac{{-3 - 7}}{{2}} = -5]

Therefore, the solutions to the equation (x^2 + 3x - 10 = 0) are (x = 2) and (x = -5).

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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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