SOLUTION: How do I solve for "x" in "-3x^2+2=29"? I know the first thing is to subtract "29" from both sides, so: "-3x^2+27=0" From there I am not certain what to do.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How do I solve for "x" in "-3x^2+2=29"? I know the first thing is to subtract "29" from both sides, so: "-3x^2+27=0" From there I am not certain what to do.      Log On


   

Question 896832: How do I solve for "x" in "-3x^2+2=29"?
I know the first thing is to subtract "29" from both sides, so:
"-3x^2+27=0"
From there I am not certain what to do.
Found 2 solutions by josgarithmetic, Alan3354:
Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
-3x%5E2=29-2
-3x%5E2=27
x%5E2=%2827%29%2F%28-3%29
x%5E2=-9
sqrt%28x%5E2%29=0%2B-+sqrt%28-9%29
highlight%28x=0%2B-+3i%29



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The 0 in the answer is only included for rendering on the page and is not really needed.
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i%5E2=-1 means that i=-sqrt%28-1%29 or i=sqrt%28-1%29.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How do I solve for "x" in "-3x^2+2=29"?
I know the first thing is to subtract "29" from both sides, so:
"-3x^2+27=0"
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-3x^2+2=29
Or, subtract 2
-3x^2 = 27
Divide by -3
x^2 = -9
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x+=+%2B3i
x+=+-3i
i+=+sqrt%28-1%29