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

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Answer 1 · 2018-06-22T16:13:19

#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!