What are the solutions to the equation (2w+1)^2=36 ?

The solutions to the equation (2w+1)^2=36 are w= 5/2 ,w=-7/2

Find the zeros of (2w+1)^2=36

The zeros of (2w+1)^2=36 are w= 5/2 ,w=-7/2

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(2 w + 1)^{2} = 36

w = \frac{5}{2}, w = - \frac{7}{2}

Decimal Notation

w = 2.5, w = - 3.5

Solution steps

(2 w + 1)^{2} = 36

For

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

the solutions are

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

Solve

2 w + 1 = 36 : w = \frac{5}{2}

Solve

2 w + 1 = - 36 : w = - \frac{7}{2}

The solutions to the quadratic equation are:

w = \frac{5}{2}, w = - \frac{7}{2}

- 11 n - 4 n = 15

\frac{x + 4}{x ^{2} - 2 x - 3} > 0

4 x + 50 = 150

\frac{2}{5} (10) + 3 (6) - 6

0 = x^{2} - 10 x + 22

What are the solutions to the equation (2w+1)^2=36 ?
The solutions to the equation (2w+1)^2=36 are w= 5/2 ,w=-7/2
Find the zeros of (2w+1)^2=36
The zeros of (2w+1)^2=36 are w= 5/2 ,w=-7/2