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

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

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

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

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

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

Decimal Notation

x = 2.5, x = - 3.5

Solution steps

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

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

(2 x + 1)^{2} = 49

s^{2} + 7 s + 11 = 0

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

(x + 3)^{2} = 15

(x + 3)^{2} = 60

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