What are the solutions to the equation (4x+4)^2-4=0 ?

The solutions to the equation (4x+4)^2-4=0 are x=-1/2 ,x=-3/2

Find the zeros of (4x+4)^2-4=0

The zeros of (4x+4)^2-4=0 are x=-1/2 ,x=-3/2

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(4 x + 4)^{2} - 4 = 0

x = - \frac{1}{2}, x = - \frac{3}{2}

Decimal Notation

x = - 0.5, x = - 1.5

Solution steps

(4 x + 4)^{2} - 4 = 0

Move

4

to the right side

(4 x + 4)^{2} = 4

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

4 x + 4 = 4 : x = - \frac{1}{2}

Solve

4 x + 4 = - 4 : x = - \frac{3}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2}, x = - \frac{3}{2}

(e^{t})^{2}

28 x^{4} + 16 x^{3} - 80 x^{2}

52 > 8 m - 4

7 x + 3 = x + 3

\frac{11}{4} - \frac{3}{2}

What are the solutions to the equation (4x+4)^2-4=0 ?
The solutions to the equation (4x+4)^2-4=0 are x=-1/2 ,x=-3/2
Find the zeros of (4x+4)^2-4=0
The zeros of (4x+4)^2-4=0 are x=-1/2 ,x=-3/2