need help

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.