algebra

First, you need to find the solutions.

3x²+5x+7=0

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-5 \pm \sqrt{5^2-4*3*7} \over 2*3}\)

x = -(5-i√59)/6, and -(5+i√59)/6

We can rewrite as:

v = -(5+i√59)/6

 u = -(5-i√59)/6

u²+v² = (-(5-i√59)/6)²+ (-(5+i√59)/6)²

= -17/9

SWest

Thx, SWest    !!!

Here's  another way

By Vieta's theorem  :

sum of  the  solutions =    -5/3  =  u + v

Square  both sides    25/9   = u^2  + 2uv + v^2      (1)

product of the  solutions  =  7/3 =  uv

So   2uv   =   14/3   

Sub this into (1)  and we  have

25/9  = u^2  + (14/3)  +  v^2

25/9  - 14/3   = u^2  + v^2

25/9   -  42/9   =  u^2  +  v^2

-17 / 9   = u^2  + v^2