What are the solutions to the equation (2x+4)^2-5=31 ?

The solutions to the equation (2x+4)^2-5=31 are x=1,x=-5

Find the zeros of (2x+4)^2-5=31

The zeros of (2x+4)^2-5=31 are x=1,x=-5

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(2 x + 4)^{2} - 5 = 31

x = 1, x = - 5

Solution steps

(2 x + 4)^{2} - 5 = 31

Move

5

to the right side

(2 x + 4)^{2} = 36

For

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

the solutions are

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

Solve

2 x + 4 = 36 : x = 1

Solve

2 x + 4 = - 36 : x = - 5

The solutions to the quadratic equation are:

x = 1, x = - 5

x^{2} - 3 (x + 2) + 8 = 0

(3 x - 4) (2 x + 3) = 0

(z + 1)^{2} = - 81

p^{2} - 14 p + 10 = - 8 p

5 x^{2} - 2 = 3 - 3 x^{2}

What are the solutions to the equation (2x+4)^2-5=31 ?
The solutions to the equation (2x+4)^2-5=31 are x=1,x=-5
Find the zeros of (2x+4)^2-5=31
The zeros of (2x+4)^2-5=31 are x=1,x=-5