What are the solutions to the equation 7(x+1)^2-28=0 ?

The solutions to the equation 7(x+1)^2-28=0 are x=1,x=-3

Find the zeros of 7(x+1)^2-28=0

The zeros of 7(x+1)^2-28=0 are x=1,x=-3

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7 (x + 1)^{2} - 28 = 0

x = 1, x = - 3

Solution steps

7 (x + 1)^{2} - 28 = 0

Move

28

to the right side

7 (x + 1)^{2} = 28

Divide both sides by

7

(x + 1)^{2} = 4

For

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

the solutions are

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

Solve

x + 1 = 4 : x = 1

Solve

x + 1 = - 4 : x = - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

0 = 3 x^{2} - 8 x + 4

a = 2 a^{2} - 1

(x + 4) (x - 4) = 20

12 x^{2} + 3 x = - 12

t^{2} - 12 t + 32 = 0

What are the solutions to the equation 7(x+1)^2-28=0 ?
The solutions to the equation 7(x+1)^2-28=0 are x=1,x=-3
Find the zeros of 7(x+1)^2-28=0
The zeros of 7(x+1)^2-28=0 are x=1,x=-3