What are the solutions to the equation 2(2x+5)^2-28=22 ?

The solutions to the equation 2(2x+5)^2-28=22 are x=0,x=-5

Find the zeros of 2(2x+5)^2-28=22

The zeros of 2(2x+5)^2-28=22 are x=0,x=-5

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2 (2 x + 5)^{2} - 28 = 22

x = 0, x = - 5

Solution steps

2 (2 x + 5)^{2} - 28 = 22

Move

28

to the right side

2 (2 x + 5)^{2} = 50

Divide both sides by

2

(2 x + 5)^{2} = 25

For

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

the solutions are

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

Solve

2 x + 5 = 25 : x = 0

Solve

2 x + 5 = - 25 : x = - 5

The solutions to the quadratic equation are:

x = 0, x = - 5

\frac{2 a}{a - 3} - \frac{2 a}{a + 3} + \frac{36}{a ^{2} - 9}

4 x^{2} - 28 x + 24

9 x^{4} + 24 x^{2} + 16

\frac{16 !}{10 !}

6 x + 175 = 3 x + 127

What are the solutions to the equation 2(2x+5)^2-28=22 ?
The solutions to the equation 2(2x+5)^2-28=22 are x=0,x=-5
Find the zeros of 2(2x+5)^2-28=22
The zeros of 2(2x+5)^2-28=22 are x=0,x=-5