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
Honestly? Just review your notes and try to work out the problems yourself before taking a look at the answer. You can also google specific topics that the test will cover and try the practice problems.