Question 442641
7y3 - 14y2 + 6y = 0

y(7y^2-14y+6)=0

y=0
Find the roots of the equation by quadratic formula						
							
a=	7	b=	-14	c=	6		
							
b^2-4ac=196-168				
b^2-4ac=28,
{{{sqrt(28)}}}=	5.29
{{{y=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{y1=(-b+sqrt(b^2-4ac))/(2a)}}}							
y1=(14+5.29)/14		
y1=1.38						
y2=(14-5.29) /	14			
y2= 0.62
...
y=0
y=1.38
y=0.62