Question 443650
{{{(4/5)y^2+(1/5)y=18/5}}}

{{{(4/5)y^2+(1/5)y-18/5=0}}}

multiply by 5
we get 4y^2+y-18=0

a=	4	b=	1	c=	-18		
							
b^2-4ac=1+288				
b^2-4ac=289
sqrt(289)=	17
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}							
x1=(-1+17)/8		
x1=2						
x2=(-1-17)/8			
x2= 	-2.25