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

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