How do I use the quadratic formula to solve #4x^2+x-1=0#?

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Answer 1 · 2014-09-30T03:46:55

#ax^2+bx+c=0#

#4x^2+1x-1=0#

#a=4, b=1, c=-1#

Quadratic formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(1)+-sqrt((1)^2-4(4)(-1)))/(2(4))#

#x=(-1+-sqrt(1+16))/8#

#x=(-1+-sqrt(17))/8#