Find the value(s) of d such that 5x^2+5(d-3)x-9d^2+15d+30=0 has two equal roots.

A quadratic equation cannot have one root. It has to have two
roots though the two may be equal. (I have made the change in the
question)

Here I guess you want the value of d so that there are two
equal roots.

Now for an equation ax^2 + bx + c = 0 to have 2 equal
roots b^2 = 4ac.

Now we
have

5X^2+5(d-3)X-9d^2+15d+30=0

a= 5 ,
b = 5(d-3) and c = -9d^2 + 15d + 30

b^2 =
4ac

=> [5(d-3)]^2 = 4(5)(-9d^2+15d +
30)

=> 25(d-3)^2 = 20( -9d^2 +15d +
30)

=> 25(d-3)^2 = 20*3(-3d^2 + 5d +
10)

=> 25(d-3)^2 = 60( -3d^2 + 5d
+10)

=> 25(d^2 - 6d + 9) = -180d^2  + 300d +
600

=> 25d^2 - 150d + 225 = -180d^2 + 300d +
600

=> 205d^2 - 450d - 375 =
0

divide by 5

=> 41d^2 - 90d -75
= 0

d1 = [-b + sqrt(b^2 -
4ac)]/2a

=> [45 + 10 sqrt 51] /
41

d2 = [-b - sqrt(b^2 -
4ac)]/2a

=> [45 - 10 sqrt 51] /
41

The values for d are [45 + 10 sqrt 51] / 41 and [45
- 10 sqrt 51] / 41