Question 177009
You can graph it and look for zeros. 
{{{ graph( 300, 300, -3, 3, -10, 10, 3x^2+x-5) }}}
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You can also calculate exact values using the quadratic formula,
{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{y = (-1 +- sqrt( 1^2-4*(3)*(-5) ))/(2*(3)) }}} 
{{{y = (-1 +- sqrt( 1+60 ))/(6) }}} 
{{{y = (-1 +- sqrt( 61))/(6) }}} 
or approximately,
{{{y[1]=(-1+sqrt(61))/6=(-1+7.81)/6=6.81/6=1.14}}}
{{{y[2]=(-1-sqrt(61))/6=(-1-7.81)/6=-8.81/6=-1.47}}}