+118724 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 :)
