What are the solutions to the equation (x-4)^2+(0+1)^2=25 ?

The solutions to the equation (x-4)^2+(0+1)^2=25 are x=2sqrt(6)+4,x=-2sqrt(6)+4

Find the zeros of (x-4)^2+(0+1)^2=25

The zeros of (x-4)^2+(0+1)^2=25 are x=2sqrt(6)+4,x=-2sqrt(6)+4

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(x - 4)^{2} + (0 + 1)^{2} = 25

x = 26 + 4, x = - 26 + 4

Decimal Notation

x = 8.89897 \dots, x = - 0.89897 \dots

Solution steps

(x - 4)^{2} + (0 + 1)^{2} = 25

Move

(0 + 1)^{2}

to the right side

(x - 4)^{2} = 24

For

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

the solutions are

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

Solve

x - 4 = 24 : x = 26 + 4

Solve

x - 4 = - 24 : x = - 26 + 4

The solutions to the quadratic equation are:

x = 26 + 4, x = - 26 + 4

[2 x - 2 (x - 1)] 5 = 4 - x

\frac{2 π}{0.6}

\frac{2 x}{3} - x = \frac{x}{15} - \frac{8}{5}

36 x^{2} - 9

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

What are the solutions to the equation (x-4)^2+(0+1)^2=25 ?
The solutions to the equation (x-4)^2+(0+1)^2=25 are x=2sqrt(6)+4,x=-2sqrt(6)+4
Find the zeros of (x-4)^2+(0+1)^2=25
The zeros of (x-4)^2+(0+1)^2=25 are x=2sqrt(6)+4,x=-2sqrt(6)+4