Complex root used to find $a$. Find $a$ if $z=2+i$ is a root of $2z^{2}+3z+2a-14+3i=0$
Workings:
$2(2+i)^{2}+3(2+i)+2a-14+3i=0$
$2(4+4i-1)+6+3i+2a-14+3i=0$
$8-8i-2+6+3i+2a-14+3i=0$
$-2+14i=-2a$
$1-7i=a$
Is this correct?

Your answer is correct.  
You could say $2a=14-3i-3z-2z^2=14-3i-3(2+i)-2(2+i)^2$
$=14-3i-3(2+i)-2(3+4i)=2-14i,$ so $a=1-7i$. 
By Vieta's formulas,  the sum of the roots is $-\dfrac32$, so the other root besides $2+i$ is $-\dfrac72-i$.  
As noted in comments, these roots are not complex conjugates; 
their product is $(2a-14+3i)/2,$ which is not real.  
