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সমাধান কর: $2\left(z^2-9\right)+9z=0$

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$2\left(z^2-9\right)+9z=0$

বা, $2z^2-18+9z=0$

বা, $2z^2+9z-18=0$

বা, $2z^2+12z-3z-18=0$

বা, $2z\left(z+6\right)-3\left(z-6\right)=0$

বা, $\left(z-6\right)\left(2z-3\right)=0$

হয়,
$\left(z-6\right)=0$

$\therefore z=6$

অথবা,
$\left(2z-3\right)=0$

বা, $2z=3$

$\therefore z=\frac32$

সতরাং নির্ণেয় সমাধান $z=6$ অথবা, $\frac32$ [Answer]

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