SOLUTION: Solve using the quadratic formula and leave the answer in radical form: y^2+6y-1=0

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Question 420394: Solve using the quadratic formula and leave the answer in radical form:
y^2+6y-1=0
Found 2 solutions by MathLover1, richard1234:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

y%5E2%2B6y-1=0....a=1, b=6 and c=-1

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

x+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+

x+=+%28-6+%2B-+sqrt%28+36%2B4+%29%29%2F2+

x+=+%28-6+%2B-+sqrt%28+40+%29%29%2F2+

x+=+%28-6+%2B-+sqrt%28+4%2A10+%29%29%2F2+

x+=+%28-6+%2B-+2%2Asqrt%28+10+%29%29%2F2+

so, you have:

x+=+%28-6+%2B+2%2Asqrt%28+10+%29%29%2F2+

x+=+-3+%2B+sqrt%28+10+%29+

or

x+=+-3+-+sqrt%28+10+%29+



Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+%28-6+%2B-+sqrt%2836-4%28-1%29%29%29%2F2 Simplify the expression.

y+=+%28-6+%2B-+sqrt%2840%29%29%2F2+=+%28-6+%2B-+2sqrt%2810%29%29%2F2+=+-3+%2B-+sqrt%2810%29