What are the solutions to the equation 3(x+1)^2+4=22 ?

The solutions to the equation 3(x+1)^2+4=22 are x=sqrt(6)-1,x=-sqrt(6)-1

Find the zeros of 3(x+1)^2+4=22

The zeros of 3(x+1)^2+4=22 are x=sqrt(6)-1,x=-sqrt(6)-1

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3 (x + 1)^{2} + 4 = 22

x = 6 - 1, x = - 6 - 1

Decimal Notation

x = 1.44948 \dots, x = - 3.44948 \dots

Solution steps

3 (x + 1)^{2} + 4 = 22

Move

4

to the right side

3 (x + 1)^{2} = 18

Divide both sides by

3

(x + 1)^{2} = 6

For

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

the solutions are

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

Solve

x + 1 = 6 : x = 6 - 1

Solve

x + 1 = - 6 : x = - 6 - 1

The solutions to the quadratic equation are:

x = 6 - 1, x = - 6 - 1

4 x + x^{2} + 12 = 0

15 x^{2} = - 5 x + 20

x^{2} - 36 x - 264 = 0

t^{2} - 3 t - 54 = 0

x (5 x + 19) = 4

What are the solutions to the equation 3(x+1)^2+4=22 ?
The solutions to the equation 3(x+1)^2+4=22 are x=sqrt(6)-1,x=-sqrt(6)-1
Find the zeros of 3(x+1)^2+4=22
The zeros of 3(x+1)^2+4=22 are x=sqrt(6)-1,x=-sqrt(6)-1