How do you solve #6x^2-x-2=0# using the quadratic formula?

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Answer 1 · 2018-05-09T08:26:00

#2/3,-1/2#

for

#ax^2+bx+c=0#

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

we ahve

#6x^2-x-2=0#

#a=6, b=-1, c=-2#

#x=(- -1+-sqrt((-1)^2-4(6)(-2)))/(2xx6)#

#x=(1+-sqrt(1+48))/12#

#x=(1+-sqrt49)/12#

#x=(1+7)/12, " or "(1-7)/12#

#x=8/12=2/3, " or "-1/2#