How do you solve #6y^2+y+-2=0#?

1 Answer
Mar 27, 2015

You simply need to solve the two equations: one with the plus sign, and the other with the minus sign.

  • Interpreting #pm# as #+#: the equation becomes #6y^2+y+2=0#. The discriminant of #6y^2+y+2# is negative, and so there are no solutions.
  • Interpreting #pm# as #-#: the equation becomes #6y^2+y-2=0#. The discriminant of #6y^2+y-2# is positive. So we can solve it, using the formula
    #y_{1,2}={-b\pm\sqrt(b^2-4ac)}/{2a}#
    Since #a=6#, #b=1# and #c=-2#, the formula becomes
    #y_{1,2}={-1\pm\sqrt(49)}/{12}={-1\pm 7}/{12}#
    So, #y_1={-1-7}/{12}=-8/12=-2/3#, and #y_2={-1+7}/{12}=6/12=1/2#