SOLUTION: Please solve the quadratic equation: x^2+12x+4=0

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Question 303397: Please solve the quadratic equation:
x^2+12x+4=0

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E2%2B12x%2B4=0 

The solution to x%5E2%2B12x%2B4=0 is

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

In your equation a=1, b=12, c=4

x+=+%28-%2812%29+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-12+%2B-+sqrt%28144-16+%29%29%2F2+

x+=+%28-12+%2B-+sqrt%28128%29%29%2F2+

x+=+%28-12+%2B-+sqrt%2864%2A2%29%29%2F2+ 

x+=+%28-12+%2B-+sqrt%2864%29sqrt%282%29%29%2F2+  

x+=+%28-12+%2B-+8%2Asqrt%282%29%29%2F2+

x+=+%284%28-3+%2B-+2%2Asqrt%282%29%29%29%2F2+

Cancel the 2 in the bottom into the 4
in the top:

x+=+2%28-3+%2B-+2%2Asqrt%282%29%29+

You can leave it like that or distribute it:

x+=+-6+%2B-+4%2Asqrt%282%29+

Edwin