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#
