What are the solutions to the equation 2(x+1)^2=40 ?

The solutions to the equation 2(x+1)^2=40 are x=2sqrt(5)-1,x=-2sqrt(5)-1

Find the zeros of 2(x+1)^2=40

The zeros of 2(x+1)^2=40 are x=2sqrt(5)-1,x=-2sqrt(5)-1

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2 (x + 1)^{2} = 40

x = 25 - 1, x = - 25 - 1

Decimal Notation

x = 3.47213 \dots, x = - 5.47213 \dots

Solution steps

2 (x + 1)^{2} = 40

Divide both sides by

2

(x + 1)^{2} = 20

For

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

the solutions are

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

Solve

x + 1 = 20 : x = 25 - 1

Solve

x + 1 = - 20 : x = - 25 - 1

The solutions to the quadratic equation are:

x = 25 - 1, x = - 25 - 1

x^{2} - 8 x - 56 = 0

x^{2} - 0.5 x - 3 = 0

4 x^{2} - 5 x = 21

24 x^{2} + 5 x - 1 = 0

4 x^{2} - 8 = 4

What are the solutions to the equation 2(x+1)^2=40 ?
The solutions to the equation 2(x+1)^2=40 are x=2sqrt(5)-1,x=-2sqrt(5)-1
Find the zeros of 2(x+1)^2=40
The zeros of 2(x+1)^2=40 are x=2sqrt(5)-1,x=-2sqrt(5)-1