Question 379480
{{{y^2=(2/3)y+(3/8)}}}
{{{y^2-(2/3)y-(3/8)=0}}}
Use the discriminant,
{{{D=b^2-4ac}}}
{{{D=(2/3)^2-4(1)(-3/8)}}}
{{{D=(4/9)+3/2}}}
{{{D=8/18+27/18}}}
{{{D=35/18}}}
Since {{{D>0}}}, there are two real and irrational roots.
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{{{graph(300,300,-5,5,-5,5,0,x^2-(2/3)x-3/8)}}}