346. Solve the equation $x^{2}+p x+q=0$ by the substitution $\boldsymbol{x}=\boldsymbol{y}+\boldsymbol{z}$ and defining $z$ (taking advantage of its arbitrariness) so as to obtain a pure quadratic equation (i.e., one not containing the first power of the unknown).
346. Given the equation:

\[
x^{2}+p x+q=0
\]

Let

\[
x=y+z
\]

Then

\[
x^{2}=y^{2}+2 y z+z^{2}
\]

Therefore, the equation becomes:

\[
\begin{aligned}
& y^{2}+2 y z+z^{2}+p y+p z+q=0 \\
& y^{2}+y(2 z+p)+z^{2}+p z+q=0
\end{aligned}
\]

154

Choose \( z \) such that the coefficient of \( y \) in the first degree disappears, i.e., \( 2 z+p \) equals zero. Then \( z=-\frac{p}{2} \). In this case, the expression:

\[
z^{2}+p z+q=\frac{p^{2}}{4}-\frac{p^{2}}{2}+q=q-\frac{p^{2}}{4}
\]

and, consequently, the equation takes the form:

\[
y^{2}=\frac{p^{2}}{4}-q
\]

from which

\[
y= \pm \sqrt{\frac{p^{2}}{4}-q}
\]

but

\[
x=y+z
\]

therefore,

\[
x=-\frac{p}{2} \pm \sqrt{\frac{p^{2}}{4}-q}
\]