How do you solve #12y^2-5y=2#?

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Answer 1 · 2016-05-20T11:46:11

The solution is:
#color(blue)(y=2/3, y=-1/4#

#12y^2 - 5y =2#

#12y^2 - 5y - 2 = 0 #

The equation is of the form #color(blue)(ay^2+by+c=0# where:

#a=12, b=-5, c=-2#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (-5)^2-(4* 12 * (-2))#

# = 25 + 96 = 121 #

The solutions are normally found using the formula
#=(-b+-sqrtDelta)/(2*a)#

#y = (-(-5)+-sqrt(121))/(2*12) = (5+-11)/24#

#y = (5 + 11 ) / 24 = 16/24 = 2/3#

#y = (5 - 11 ) / 24 = -6/24 = -1/4#