What are the solutions to the equation (5x+n)^2-4=0 ?

The solutions to the equation (5x+n)^2-4=0 are x=(2-n)/5 ,x=(-2-n)/5

Find the zeros of (5x+n)^2-4=0

The zeros of (5x+n)^2-4=0 are x=(2-n)/5 ,x=(-2-n)/5

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(5 x + n)^{2} - 4 = 0

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

Solution steps

(5 x + n)^{2} - 4 = 0

Move

4

to the right side

(5 x + n)^{2} = 4

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

e (7) = - x^{2} - 9 x

x^{2} = 1 7^{2} + 8^{2}

(3 x - 1) (2 x + 3) = x + 2 (x - 1)

5 x^{2} + 10 x = x - 4

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

What are the solutions to the equation (5x+n)^2-4=0 ?
The solutions to the equation (5x+n)^2-4=0 are x=(2-n)/5 ,x=(-2-n)/5
Find the zeros of (5x+n)^2-4=0
The zeros of (5x+n)^2-4=0 are x=(2-n)/5 ,x=(-2-n)/5