Suppose $2+7i$ is a solution of $2z^2+Az+B=0$, where $A,B \in \mathbb{R}$ . Find $A$ and $B$ The question is as follows;

Suppose  $2+7i$  is a solution of  $2z^2+Az+B=0$, where  $A, B \in \mathbb{R}$ . Find $A$ and $B$.

My understanding is that this equation holds:
$$2(2+7i)^2 + A(2+7i) + B = 0$$
which will eventually lead to:
$$-90 + 2A + B + i(56+7A) = 0$$
I would like to check if my approach is correct, and if so, what should I do next to derive $A$ & $B$.

A polynomial with real coefficients will have non-real roots in conjugate pairs. Since it has $2+7i$ as a root, it will also have $2-7i$ as a root. Being a degree $2$ polynomial, these are the only roots. The leading coefficient of $2z^2+Az+B=0$ is $2$, so our polynomial is:
$$2(z- (2-7i))(z -(2+7i)) = 0$$
Expanding, the coefficients we obtain:
$$2 z^2 - 2(2-7i + 2+7i)z + 2(2-7i)(2+7i)  = 0\\
2z^2 - 8z + 106 = 0$$
We get:  $A=-8$, $B = 106$.
