SOLUTION: I cannot figure out how to solve this quadratic equation. Solve for x: 2x^2 + 3bx + b^2 = 0 Thank you, Melody Salzer

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Question 88446: I cannot figure out how to solve this quadratic equation.
Solve for x:
2x^2 + 3bx + b^2 = 0
Thank you,
Melody Salzer
Found 3 solutions by stanbon, Earlsdon, checkley75:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
2x^2 + 3bx + b^2 = 0
------------
Use the quadratic formula to get:
x = [-3b+-sqrt(9b^2-4*2*b^2)]/4
x = [-3b +-sqrt(b^2)]/4
x = [-3b+-b]/4
x = -b or x = -b/2
==================
Cheers,
Stan H.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you can always use the quadratic formula to solve a quadratic equation: x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
But just to make it a little less confusing, we'll write your equation like this, with a B instead of a b:
2x%5E2%2B3Bx%2BB%5E2+=+0 so...a = 2, b = 3B, and c = B^2
Now, making the appropriate substitutions, we get:
x+=+%28-3B%2B-sqrt%28%283B%29%5E2-4%282%29%28B%5E2%29%29%29%2F2%282%29 Simplifying this, we have:
x+=+%28-3B%2B-sqrt%289B%5E2-8B%5E2%29%29%2F4
x+=+%28-3B%2B-sqrt%28b%5E2%29%29%2F4
x+=+%28-3B%2B-B%29%2F4
So the roots are:
x+=+%28-3B%2BB%29%2F4 or x+=+-3B-B%29%2F4
x+=+-2B%2F4 or x+=+-4B%2F4
x+=+-B%2F2 or x+=+-B
But remember that we substituted the b in the original equation with B, so the answers are:
x+=+-b%2F2 or x+=+-b

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2+3bx+b^2=0
using the quadratic equation we get:
x=(-3b+-sqrt[(3b)^2-4*2*b^2])/2*2
x=-3b+-sqrt[9b^2-8b^2])/4
x=(-3b+-sqrtb^2)/4
x=(-3b+-b)/4
x=(-3b+b)/4
x=-2b/4
x=-b/2 answer
x=(-3b-b)/4
x=-4b/4
x=-b answer.