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Pre-Algebra
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I am asked to solve 42=x^2-x How do I solve this?
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Notice how if you subtract 42 from both sides you get \[x^2 - x - 42 = 0\] Does that look like something familiar?
So in order to solve this, you need to set the right side equal to 0 by subtracting 42. \[x ^{2}-x-42\] Next you need to factor this. The factors of -42 that equal -1 are 6 and -7, so (x+6)(x-7)
*\[x ^{2}-x-42=0\]
Got it thanks for the help
You're very welcome :)
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