What are the solutions to the equation (2x+1)^2=9 ?

The solutions to the equation (2x+1)^2=9 are x=1,x=-2

Find the zeros of (2x+1)^2=9

The zeros of (2x+1)^2=9 are x=1,x=-2

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(2 x + 1)^{2} = 9

x = 1, x = - 2

Solution steps

(2 x + 1)^{2} = 9

For

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

the solutions are

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

Solve

2 x + 1 = 9 : x = 1

Solve

2 x + 1 = - 9 : x = - 2

The solutions to the quadratic equation are:

x = 1, x = - 2

10 \cdot \frac{1}{5}

5 (0) + 7 (2) \geq - 3

- x^{2} - 8 = - 33

1 0^{2} = r^{2} + r^{2} - 2 \cdot r \cdot r \cdot \frac{1 + 5}{4}

\frac{5 c ^{2} d ^{3}}{- 4 c ^{2} d}

What are the solutions to the equation (2x+1)^2=9 ?
The solutions to the equation (2x+1)^2=9 are x=1,x=-2
Find the zeros of (2x+1)^2=9
The zeros of (2x+1)^2=9 are x=1,x=-2