What are the solutions to the equation (2x+1)^2=16 ?

The solutions to the equation (2x+1)^2=16 are x= 3/2 ,x=-5/2

Find the zeros of (2x+1)^2=16

The zeros of (2x+1)^2=16 are x= 3/2 ,x=-5/2

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(2 x + 1)^{2} = 16

x = \frac{3}{2}, x = - \frac{5}{2}

Decimal Notation

x = 1.5, x = - 2.5

Solution steps

(2 x + 1)^{2} = 16

For

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

the solutions are

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

Solve

2 x + 1 = 16 : x = \frac{3}{2}

Solve

2 x + 1 = - 16 : x = - \frac{5}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - \frac{5}{2}

a x^{2} - c = 0

x, 2 x^{2} + 2 y^{2} + 10 = 0

(x - 5)^{2} = 25 - 9 x

x, \frac{x ^{2}}{3} + 3 y^{2} = - 3 + 3

320 - 320 x - x^{2} = 0

What are the solutions to the equation (2x+1)^2=16 ?
The solutions to the equation (2x+1)^2=16 are x= 3/2 ,x=-5/2
Find the zeros of (2x+1)^2=16
The zeros of (2x+1)^2=16 are x= 3/2 ,x=-5/2