Need to know the nature of its roots.

d2 + 7/3d = 2

Need to find the nature of its roots.

$d^2+\frac{7}{3}d=2\\ d^2+\frac{7}{3}d-2=0\\ \mbox{It is easiest to multiply this by 3 to get rid of the fraction}\\ 3d^2+7d-6=0\\ $

Now the nature of the roots is determined by the discriminant.  That it the bit under the square root in the quadratic formula.

The symbol for the discriminant is a triangle.

$\triangle=b^2-4ac\\ \triangle=7^2-4*3*-6\\ \triangle=49+72\\ \triangle=121\\ $

Now 121 is a perfect square so this means there will be two rational roots :)

If you want more explanation then ask for it :)