x^{2} + 2(m – 1)x + (m + 5) = 0
Equating with ax^{2} + bx + c = 0
a = 1, b = 2(m – 1), c = (m + 5)
Since equation has real and equal roots.
So, D = 0
⇒ b^{2} – 4ac = 0
[2(m – 1)^{2} – 4 × 1 × (m + 5) = 0
⇒ 4(m – 1)^{2} – 4(m + 5) = 0
⇒ 4 [(m – 1)^{2} – (m + 5)] = 0
⇒ 4 [m^{2} – 2m + 1 – m – 5] = 0
⇒ m^{2} – 3m – 4 = 0
⇒ (m + 1)(m – 4) = 0
Either m + 1 = 0
m = - 1
or m – 4 = 0
m = 4
m = -1, 4