What are the solutions to the equation (x+4)^2=23 ?

The solutions to the equation (x+4)^2=23 are x=sqrt(23)-4,x=-sqrt(23)-4

Find the zeros of (x+4)^2=23

The zeros of (x+4)^2=23 are x=sqrt(23)-4,x=-sqrt(23)-4

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(x + 4)^{2} = 23

x = 23 - 4, x = - 23 - 4

Decimal Notation

x = 0.79583 \dots, x = - 8.79583 \dots

Solution steps

(x + 4)^{2} = 23

For

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

the solutions are

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

Solve

x + 4 = 23 : x = 23 - 4

Solve

x + 4 = - 23 : x = - 23 - 4

The solutions to the quadratic equation are:

x = 23 - 4, x = - 23 - 4

7 x + 31 = 4 x - 11

∣ 9 x - 4 ∣ + 3 = 5

x^{2} - 2 x - 7 > 1

- 10 = x - 21

\frac{4 + x}{2} - \frac{x - 3}{5} < - \frac{x}{10}

What are the solutions to the equation (x+4)^2=23 ?
The solutions to the equation (x+4)^2=23 are x=sqrt(23)-4,x=-sqrt(23)-4
Find the zeros of (x+4)^2=23
The zeros of (x+4)^2=23 are x=sqrt(23)-4,x=-sqrt(23)-4