How do you solve for y in #6y^2-5y=6#?

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2 Answers
Candy
May 25, 2018

#color(blue)(y=3/2, -2/3#

Explanation:

#6y^2-5y=6#

#6y^2-5y-6=0#

#6y^2-9y+4y-6=0#

#3y(2y-3)+2(2y-3)=0#

#(2y-3)(3y+2)=0#

#therefore 2y-3=0 or 3y+2=0#

#thus, y=3/2, -2/3#

Do not forget to check your answer:
Plug in #color(teal)(y=3/2# and #color(teal)(y=-2/3# in the equation.

#6(3/2)^2-5(3/2)=6*9/4-15/2=27/2-15/2=12/2=6#
That proves that #color(teal)(y=3/2# is correct

Next, check for #-2/3:#
#6(-2/3)^2-5(-2/3)=6*4/9+10/3=8/3+10/3=18/3=6#

That's it your final answer is:
#color(blue)(y=3/2, -2/3#

Nallasivam V
May 25, 2018

#y=2/3#
#y=-3/2#

Explanation:

Given -

#6y^2-5y=6#
divide both sides by 6

#y^2-5/6y=1#

#y^2-5/6y+25/144=1+25/144=(144+25)/144=169/144#

#(y+5/12)^2=169/144#

#y+5/12=+-sqrt(169/144)=+-13/12#

#y=13/12-5/12=8/12=2/3#

#y=-13/12-5/12=-18/12=-3/2#

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