Question

x^2-(2b - 1) x + (b^2 -b-20)= 0

Answer 1

Hey friend, Harish here.

Here is your answer:

⇒ $x^{2}-(2b - 1) x + (b^{2} -b-20)= 0$

⇒ $x^{2}-(2b - 1) x + (b^{2} +4b -5b-20)= 0$

⇒ $x^{2}-(2b - 1) x + ((b+4)(b-5))= 0$

⇒ $x^2 -(b+4)x-(b-5)x + ((b+4)(b-5))= 0$

⇒  $x(x-b-4)-(b-5)(x-b-4)=0$

⇒  $(x-b+5)(x-b-4)=0$

Then, x = b - 5, & x = b + 4
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Hope my answer is helpful to you.