Divide \(3x^2+4x-12=0\) by 3:  \(x^2+\frac{4}{3}x-4=0\)   ...(1)
  
 Quadratic with roots r and s:  \((x-r)(x-s)=0\)  or  \(x^2-(r+s)x+rs=0\)    ...(2)
  
 Compare (1) and (2)  to see that \(r+s=-\frac{4}{3}\)    ...(3)
  
 Note that \(r^2 +s^2 + 2rs = (r+s)^2 \)
  
 Now it should be easy to answer the question.