simple math question

Hello Konzetsu!

I mistakenly entered at the end of x instead of p.
Please indulgence! Sorry!

16 - 3p = 2/3p + 5

$16 - 3p = 2/3p + 5$

$11 = 3p + 2/3p$

$11 = \frac{9\ p^{2}+ 2 }{3p}$

$33p = 9p^{2}+ 2$

$9p^{2}- 33p+ 2= 0$

$a= 9; b= -33;c= 2$

$p = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$p = {33 \pm \sqrt{33^2-4*9*2} \over 2*9}$  

$p_{1}= \frac{33+ \sqrt{1017} }{18} = 3.605$

$p_{2}= \frac{33- \sqrt{1017} }{18} = 0.06164$

Greeting asinus :- )  !