SOLUTION: Find the zeros (x-intercepts): X^2-6x-5=2 Please explain how you solve these type of equations because I have a important test coming up and I really need help with solving thes

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Question 1008667: Find the zeros (x-intercepts):
X^2-6x-5=2
Please explain how you solve these type of equations because I have a important test coming up and I really need help with solving these types of equations! Thanks so much!
Found 2 solutions by fractalier, Fombitz:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
The zeroes of a function mean what are the x-values that make the graph of the function cross the x-axis. Remember that when a line crosses the x-axis, its y-value is zero.
So we set these functions equal to zero and solve for x.
Here we have
x^2 - 6x - 5 = 2 or
x^2 - 6x - 7 = 0
Now we can factor this and get (we do this by un-FOILing)
(x - 7)(x + 1) = 0
and thus
x = 7 and x = -1
Does that help?

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Factoring:
x%5E2-6x-5=2
x%5E2-6x-7=0
Factor.
%28x-7%29%28x%2B1%29=0
Two solutions:
x-7=0
x=7
and
x%2B1=0
x=-1
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.
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You could also have used the quadratic formula,
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
with
a=1
b=-6
c=-7
x+=+%286+%2B-+sqrt%28+36-4%2A1%2A%28-7%29+%29%29%2F%282%2A1%29+
x+=+%286+%2B-+sqrt%28+36%2B28+%29%29%2F%282%29+
x+=+%286+%2B-+sqrt%28+64%29%29%2F%282%29+
x+=+%286+%2B-+8%29%2F%282%29+
x+=+3+%2B-+4
x=7 and x=-1
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.
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You could also graph to find the roots although this is the least precise method.
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