solve for x. 4x^2+4x+1=49?

4x²+4x+1=49
There are two ways to solve this.
The first is simply using the property of squares
4x²+4x+1=49
(2x + 1)² = 49
2x + 1 = ± 7
2x = 1 ± 7
2x = -6 or 8
x = -3 or 4

The other way is to solve a quadratic equation
4x²+4x+1=49
4x²+4x- 48 = 0............Subtract 49 from bit sides
x² + x - 12 = 0.............Divide by 4
(x + 3)(x - 4) = 0..........Factor
You will get the same answers:
x = -3 or x = 4

Select your method.

4x^2 + 4x + 1 = 49
4x^2 + 2x + 2x + 1 = 7^2
(4x^2 + 2x) + (2x + 1) = 7^2
2x(2x + 1) + 1(2x + 1) = 7^2
(2x + 1)(2x + 1) = 7^2
(2x + 1)^2 = 7^2
2x + 1 = 7
2x = 7 - 1
x = 6/2
x = 3

Put it equal to zero:
4x² + 4x + 1 = 49
4x² + 4x - 48 = 0

Divide it by 4 to make it easier to factorise:
4(x² + x - 12) = 0

Factorise that:
4(x + 4)(x - 3) = 0

x = -4
x = 3

Not sure how that is related to a perfect square?

4x² + 4x - 48 = 0
x² + x - 12 = 0
(x + 4)(x - 3) = 0
x = - 4 , x = 3