What are the solutions to the equation (2n+3)^2=4 ?

The solutions to the equation (2n+3)^2=4 are n=-1/2 ,n=-5/2

Find the zeros of (2n+3)^2=4

The zeros of (2n+3)^2=4 are n=-1/2 ,n=-5/2

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(2 n + 3)^{2} = 4

n = - \frac{1}{2}, n = - \frac{5}{2}

Decimal Notation

n = - 0.5, n = - 2.5

Solution steps

(2 n + 3)^{2} = 4

For

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

the solutions are

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

Solve

2 n + 3 = 4 : n = - \frac{1}{2}

Solve

2 n + 3 = - 4 : n = - \frac{5}{2}

The solutions to the quadratic equation are:

n = - \frac{1}{2}, n = - \frac{5}{2}

(x) = 4 x^{2} - 3 x + 25

z^{2} + 6 z - 8 = 0

74 - 2 x^{2} = 84

9 x^{2} - 5 x (\frac{28 x}{33}) + 7 (\frac{28 x}{33})^{2} = 120

(x + 5) x = 0

What are the solutions to the equation (2n+3)^2=4 ?
The solutions to the equation (2n+3)^2=4 are n=-1/2 ,n=-5/2
Find the zeros of (2n+3)^2=4
The zeros of (2n+3)^2=4 are n=-1/2 ,n=-5/2