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

The solutions to the equation (x-4)^2+25=0 are x=5i+4,x=-5i+4

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

The zeros of (x-4)^2+25=0 are x=5i+4,x=-5i+4

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

x = 5 i + 4, x = - 5 i + 4

Solution steps

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

Move

25

to the right side

(x - 4)^{2} = - 25

For

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

the solutions are

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

Solve

x - 4 = - 25 : x = 5 i + 4

Solve

x - 4 = - - 25 : x = - 5 i + 4

The solutions to the quadratic equation are:

x = 5 i + 4, x = - 5 i + 4

t^{2} + 7 t + 12 = 0

- 1 + \frac{1}{2}

5 x^{2} - 7 x = 2

\frac{42 - 62}{62}

\frac{1}{y ^{\frac{5}{4}}}

What are the solutions to the equation (x-4)^2+25=0 ?
The solutions to the equation (x-4)^2+25=0 are x=5i+4,x=-5i+4
Find the zeros of (x-4)^2+25=0
The zeros of (x-4)^2+25=0 are x=5i+4,x=-5i+4