What are the solutions to the equation (d-1)^2=-18 ?

The solutions to the equation (d-1)^2=-18 are d=3sqrt(2)i+1,d=-3sqrt(2)i+1

Find the zeros of (d-1)^2=-18

The zeros of (d-1)^2=-18 are d=3sqrt(2)i+1,d=-3sqrt(2)i+1

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(d - 1)^{2} = - 18

d = 32 i + 1, d = - 32 i + 1

Solution steps

(d - 1)^{2} = - 18

For

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

the solutions are

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

Solve

d - 1 = - 18 : d = 32 i + 1

Solve

d - 1 = - - 18 : d = - 32 i + 1

The solutions to the quadratic equation are:

d = 32 i + 1, d = - 32 i + 1

21 6^{\frac{4}{3}}

- \frac{9}{10} p - 3 = \frac{3}{5}

\frac{1}{x} + \frac{1}{x + 1} < \frac{2}{x + 9}

11 \geq 3 x + 2 and 20 - 3 (x + 2) \leq 5

v^{2} + 21 = - 8 v + 6

What are the solutions to the equation (d-1)^2=-18 ?
The solutions to the equation (d-1)^2=-18 are d=3sqrt(2)i+1,d=-3sqrt(2)i+1
Find the zeros of (d-1)^2=-18
The zeros of (d-1)^2=-18 are d=3sqrt(2)i+1,d=-3sqrt(2)i+1