What are the solutions to the equation 3(x+1)^2-4=5 ?

The solutions to the equation 3(x+1)^2-4=5 are x=sqrt(3)-1,x=-sqrt(3)-1

Find the zeros of 3(x+1)^2-4=5

The zeros of 3(x+1)^2-4=5 are x=sqrt(3)-1,x=-sqrt(3)-1

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3 (x + 1)^{2} - 4 = 5

x = 3 - 1, x = - 3 - 1

Decimal Notation

x = 0.73205 \dots, x = - 2.73205 \dots

Solution steps

3 (x + 1)^{2} - 4 = 5

Move

4

to the right side

3 (x + 1)^{2} = 9

Divide both sides by

3

(x + 1)^{2} = 3

For

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

the solutions are

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

Solve

x + 1 = 3 : x = 3 - 1

Solve

x + 1 = - 3 : x = - 3 - 1

The solutions to the quadratic equation are:

x = 3 - 1, x = - 3 - 1

8 x^{2} - 24 x = 0

x^{2} + 5 x - 150 = 0

a^{2} - 4 a + 4 = 0

(y + 1) (y + 3) = - 2

4 (3 x - 7)^{2} = 49

What are the solutions to the equation 3(x+1)^2-4=5 ?
The solutions to the equation 3(x+1)^2-4=5 are x=sqrt(3)-1,x=-sqrt(3)-1
Find the zeros of 3(x+1)^2-4=5
The zeros of 3(x+1)^2-4=5 are x=sqrt(3)-1,x=-sqrt(3)-1