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

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

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

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

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

x = 10, x = - 2

Solution steps

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

Switch sides

(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 = 10

Solve

x - 4 = - 36 : x = - 2

The solutions to the quadratic equation are:

x = 10, x = - 2

3 x^{2} y + 4 x y^{2} \cdot x - y

(3 + 4 i) \cdot (3.2 + 1.4 i)

7 x^{2} + 2 x = 2

2 x^{2} - 8 x + 9 = 0

2 x \leq 10

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