Algebra Help! Im stuck

Let r and s be the roots of 3x^2+4x+12=0. Find r^2+s^2.

Hello Age2!

\(3x^2+4x+12=0\)

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

\(x = {-2 \pm \sqrt{16-4\cdot 3\cdot 12} \over 2\cdot 3}\)

The parabola has no real solution.

\(x=-\frac{2}{6}\pm \frac{i}{6}\cdot \sqrt{128}\\ x=-\frac{2}{6}\pm \frac{i}{6}\cdot \sqrt{2\cdot 2^6}\\ x=-\frac{2}{6}\pm \frac{8i}{6}\cdot \sqrt{2}\\ \color{blue}x=-\frac{1}{3}\pm \frac{4}{3}\cdot \sqrt{2}\cdot i\\ \)

\(r=-\frac{1}{3}+ \frac{4}{3}\cdot \sqrt{2}\cdot i\\ s=-\frac{1}{3}- \frac{4}{3}\cdot \sqrt{2}\cdot i \)

\(r=-\frac{1}{3}+ \frac{4}{3}\cdot \sqrt{2}\cdot i\\ s=-\frac{1}{3}- \frac{4}{3}\cdot \sqrt{2}\cdot i \)

\(r^2=\frac{1}{9}-\frac{8}{9}\cdot \sqrt{2}\cdot i+\frac{16}{9}\cdot 2\cdot 1\\ s^2=\frac{1}{9}+\frac{8}{9}\cdot \sqrt{2}\cdot i+\frac{16}{9}\cdot 2\cdot 1\)

to be continued

  !