What are the solutions to the equation (2x+3)^2=1 ?

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

Find the zeros of (2x+3)^2=1

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

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(2 x + 3)^{2} = 1

x = - 1, x = - 2

Solution steps

(2 x + 3)^{2} = 1

For

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

the solutions are

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

Solve

2 x + 3 = 1 : x = - 1

Solve

2 x + 3 = - 1 : x = - 2

The solutions to the quadratic equation are:

x = - 1, x = - 2

9 x^{2} = 11 + 6 x

x, 2 x^{2} h = 1000

A (x) = (x - 5) (x + 7)

3 x^{2} + 660 x - 15 = 0

6 y^{2} + 3 y + 24 = 0

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