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Question 138232: How do you solve for each variable:
(x-3)(x-7)=0
b^2+3b-4=0
n^2+n-12=0
5x^2+18x=8
3a^2+4a=2a^2-2a-9
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (x-3)(x-7)=0
the only way the product can be zero is if one of the factors is zero.
x = 3 or x = 7
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b^2+3b-4=0
(b+4)(b-1)=0
b = -4 or b = 1
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n^2+n-12=0
(n+4)(n-3)=0
n=-4 or n=3
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5x^2+18x=8
5x^2+18x-8 = 0
(5x-2)(x+4) = 0
x = 2/5 or x = -4
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3a^2+4a=2a^2-2a-9
a^2+6a+9 = 0
(a+3)(a+3) = 0
a=-3 with multiplicity two
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Cheers,
Stan H.
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