Determine the value of variable:  (x+5)^2-4(4x+5)=0

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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

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giorgiana1976 | Student

To solve the equation, we'll expand the square and we'll remove the brackets.

We'll expand the square using the formula:

(a+b)^2 = a^2 + 2ab + b^2

Now, we'll expand (x+5)^2:

(x+5)^2 = x^2 + 10x + 25

The equation will become:

x^2 + 10x + 25 - 16x - 20 = 0

We'll combine like terms:

x^2 - 6x + 5 = 0

We'll apply quadratic formula;

x1 = [6+sqrt(36 - 20)]/2

x1 = (6+4)/2

x1 = 5

x2 = (6-4)/2

x2 = 1

The solutions of the equation are {1 ; 5}.

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