We have to solve (x+5)^2-4(4x+5)=0

(x+5)^2 - 4(4x+5) = 0

open the brackets

x^2 + 10x + 25 - 16x - 20 = 0

=> x^2 - 6x + 5 = 0

=> x^2 - 5x - x + 5 = 0

=> x( x - 5) - 1(x - 5) = 0

=> (x - )(x - 5) = 0

**Therefore x = 1 and x = 5**

We have to solve (x+5)^2-4(4x+5)=0

(x+5)^2 - 4(4x+5) = 0

open the brackets

x^2 + 10x + 25 - 16x - 20 = 0

=> x^2 - 6x + 5 = 0

=> x^2 - 5x - x + 5 = 0

=> x( x - 5) - 1(x - 5) = 0

=> (x - )(x - 5) = 0

**Therefore x = 1 and x = 5**