Question

Hey need some help quick

Answer 1

If $z=\pm 15i$ has to be a solution of the equation, just plug it in:

$(\pm 15i)^2 + [1] = 0 \iff -225 + [1] = 0 \iff [1] = 225$

So, the solutions of

$z^2+225=0$

are $z=\pm 15i$