How do you solve #10+a^2=-7a#?

Answer 1

#x=-2# and #x=-5#

Since we have a second-degree term, we know we are dealing with a quadratic, so we need to set it equal to zero to find its zeroes.

We can add #7a# to both sides to get
#a^2+7a+10=0#
At this point, we want to think of two numbers that sum up to the middle term (#7#) and have a product of the last term (#10#).
After some trial and error, we arrive at #5# and #2#. This means we can factor this as
#(x+5)(x+2)=0#

Setting both factors equal to zero, we get

#x=-2# and #x=-5#

Hope this helps!

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Answer 2

To solve (10 + a^2 = -7a), first, bring all terms to one side to set the equation to zero: (a^2 + 7a + 10 = 0). Then, factor the quadratic equation: ((a + 5)(a + 2) = 0). Finally, solve for (a): (a = -5) or (a = -2).

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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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