What are the solutions to the equation 3(x+1)^2-1=17 ?

The solutions to the equation 3(x+1)^2-1=17 are x=sqrt(6)-1,x=-sqrt(6)-1

Find the zeros of 3(x+1)^2-1=17

The zeros of 3(x+1)^2-1=17 are x=sqrt(6)-1,x=-sqrt(6)-1

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3 (x + 1)^{2} - 1 = 17

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

Decimal Notation

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

Solution steps

3 (x + 1)^{2} - 1 = 17

Move

1

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

5 x + 4 = 14

det (\frac{4}{2} \frac{- 3}{- 5})

lo g_{10} (\frac{5}{7})

2 x^{3} - 3 x^{2} - 8 x - 3

\frac{3 i ^{2} + 3 i}{4 i}

What are the solutions to the equation 3(x+1)^2-1=17 ?
The solutions to the equation 3(x+1)^2-1=17 are x=sqrt(6)-1,x=-sqrt(6)-1
Find the zeros of 3(x+1)^2-1=17
The zeros of 3(x+1)^2-1=17 are x=sqrt(6)-1,x=-sqrt(6)-1