What are the solutions to the equation (x+1)^2+4=14 ?

The solutions to the equation (x+1)^2+4=14 are x=sqrt(10)-1,x=-sqrt(10)-1

Find the zeros of (x+1)^2+4=14

The zeros of (x+1)^2+4=14 are x=sqrt(10)-1,x=-sqrt(10)-1

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(x + 1)^{2} + 4 = 14

x = 10 - 1, x = - 10 - 1

Decimal Notation

x = 2.16227 \dots, x = - 4.16227 \dots

Solution steps

(x + 1)^{2} + 4 = 14

Move

4

to the right side

(x + 1)^{2} = 10

For

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

the solutions are

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

Solve

x + 1 = 10 : x = 10 - 1

Solve

x + 1 = - 10 : x = - 10 - 1

The solutions to the quadratic equation are:

x = 10 - 1, x = - 10 - 1

- \frac{1}{6} (2 u - 5) = 4 (- 2 u + 5)

x^{2} - 21 x + 90

\frac{3 ^{8}}{3 ^{6}}

- x^{4} + 17 x^{2} - 16

4 x^{2} - 8 x - 5

What are the solutions to the equation (x+1)^2+4=14 ?
The solutions to the equation (x+1)^2+4=14 are x=sqrt(10)-1,x=-sqrt(10)-1
Find the zeros of (x+1)^2+4=14
The zeros of (x+1)^2+4=14 are x=sqrt(10)-1,x=-sqrt(10)-1