What are the solutions to the equation 2(2x+1)^2-2=160 ?

The solutions to the equation 2(2x+1)^2-2=160 are x=4,x=-5

Find the zeros of 2(2x+1)^2-2=160

The zeros of 2(2x+1)^2-2=160 are x=4,x=-5

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2 (2 x + 1)^{2} - 2 = 160

x = 4, x = - 5

Solution steps

2 (2 x + 1)^{2} - 2 = 160

Move

2

to the right side

2 (2 x + 1)^{2} = 162

Divide both sides by

2

(2 x + 1)^{2} = 81

For

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

the solutions are

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

Solve

2 x + 1 = 81 : x = 4

Solve

2 x + 1 = - 81 : x = - 5

The solutions to the quadratic equation are:

x = 4, x = - 5

x^{2} + 4 x + 21 = 0

\frac{2}{10}

- \frac{8 + r}{2} > - 8

(5 x - 3)^{2} = 25

3 x^{2} - 11 = 0

What are the solutions to the equation 2(2x+1)^2-2=160 ?
The solutions to the equation 2(2x+1)^2-2=160 are x=4,x=-5
Find the zeros of 2(2x+1)^2-2=160
The zeros of 2(2x+1)^2-2=160 are x=4,x=-5