# How do you solve #x^2 + 5x + 1 = 0# using the quadratic formula?

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To solve the quadratic equation ( x^2 + 5x + 1 = 0 ) using the quadratic formula, ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ), where ( a = 1 ), ( b = 5 ), and ( c = 1 ).

Substitute the values of ( a ), ( b ), and ( c ) into the formula: ( x = \frac{{-5 \pm \sqrt{{5^2 - 4 \cdot 1 \cdot 1}}}}{{2 \cdot 1}} )

Calculate the discriminant: ( \text{Discriminant} = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 1 = 25 - 4 = 21 )

Apply the square root to the discriminant: ( \sqrt{21} )

Then, the solutions are: ( x = \frac{{-5 + \sqrt{21}}}{{2}} ) and ( x = \frac{{-5 - \sqrt{21}}}{{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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