What are the solutions to the equation (x+4)^2-5=31 ?

The solutions to the equation (x+4)^2-5=31 are x=2,x=-10

Find the zeros of (x+4)^2-5=31

The zeros of (x+4)^2-5=31 are x=2,x=-10

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(x + 4)^{2} - 5 = 31

x = 2, x = - 10

Solution steps

(x + 4)^{2} - 5 = 31

Move

5

to the right side

(x + 4)^{2} = 36

For

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

the solutions are

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

Solve

x + 4 = 36 : x = 2

Solve

x + 4 = - 36 : x = - 10

The solutions to the quadratic equation are:

x = 2, x = - 10

x^{2} + 4 x - 8 = 0

4 x^{2} + 16 x

m, F = \frac{g mm}{d ^{2}}

1 0^{2} + b^{2} = 1 5^{2}

x, x^{2} + (y - 1)^{2} = 0

What are the solutions to the equation (x+4)^2-5=31 ?
The solutions to the equation (x+4)^2-5=31 are x=2,x=-10
Find the zeros of (x+4)^2-5=31
The zeros of (x+4)^2-5=31 are x=2,x=-10