y^2-5/3 y =-1/3 solve in quadratic formula
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y^2-5/3y= -1/3
3y^2-5y/3 =-1/3
3y^2-5y=-1
3y^2-5y+1=0
Now,by Shridharacharya formula,we have
a=3,,b=5 and c=1 so,
D=(b^2-4ac)
=(5^2-4.3.1)
=(25-12)=13..
y=-b(+-)√d/2a. (since,+-means +or-).
y=-5(+-)√13/2.3
y=-5(+-)√13/6
Now,there is two answer in + and -..
then,y=-5+√13/6 and y=-5-√13/6...
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