How do you solve #6x²=1-x#?

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Answer 1 · 2016-10-04T20:34:17

#" at "y=0 :" "x=-1/2" and "x=1/3#

Write as #" "y=6x^2+x-1#

Compare to the form #" "y=ax^2+bx+c#

#color(blue)("Using the formula: "x=(-b+-sqrt(b^2-4ac))/(2a))#

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

#x=(-1+-sqrt(25))/12" "->" "x=-1/12+-5/12#

#=>" at "y=0 :" "x=-1/2" and "x=1/3#

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Tony B