Given the roots of both the equations are real

For first equation ax² + 2bx + c = 0

Its discriminant; d ≥ 0

D = b² – 4ac

D = (2b)² – 4 × a × c ≥ 0

4b² ≥

b² ≥ ac …1

For second equation

bx² - 2x + b = 0

d = b² – 4ac ≥ 0

= (2)² – 4 b xb ≥ 0

= 4ac – 4 b² ≥

ac ≥ b² …2

From 1 and 2 we get only one case where b² = ac