so we see the leading coefient is positive (the 3 in the 3x^2 is positive) so the paraobola opens upward
so for it to have 2 x intercepts, we must have the vertex below the x axis
we must have the y value of the vertex negative
so
a hack is this
to move a function up c units, add c to whole function
also
for
y=ax^2+bx+c
the x value of the vertex is -b/2a
the y value is found by subsituting that value for x
so
y=3x^2+7x+m
x value of vertex is
-7/(2*3)=-7/6
if we sub it in
y=3(-7/6)^2+7(-7/6)+m
y=49/12-49/6+m
we want the y value to be less than 0
so
0>49/12-49/6+m
0>49/12-98/12+m
0>-49/12+m
49/12>m
so it will have 2 x intercept for all m such that m<49/12