How do you solve #3r ^ { 2} - 24= 6r#?

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Answer 1 · 2018-06-30T21:05:23

#x=4# and #x=-2#

We are dealing with a quadratic, so we want to set it equal to zero to find its factors. Let's subtract #6x# from both sides to get

#3r^2-6r-24=0#, which is the same as

All terms have a #3# in common, so we can factor that out. We get

#3(r^2-2r-8)=0#

Let's pay attention to what's in the parenthesis. What two numbers sum up to #-2# and have a product of #-8#?

After some trial and error, we arrive at

#-4# and #2#. This means we can factor this as

#3(x-4)(x+2)=0#

Setting both factors equal to zero, we get

#x=4# and #x=-2#

Hope this helps!