Roots squared

By Vieta's formula , $x_{1} + x_{2} = \dfrac{3}{2}$ and $x_{1}x_{2} = \dfrac{9}{2}$ .

Thus $x_{1}^{2} + x_{2}^{2} = (x_{1} + x_{2})^{2} - 2x_{1}x_{2} = \left(\dfrac{3}{2}\right)^{2} - 2 \times \dfrac{9}{2} = \boxed{-\dfrac{27}{4}}$ .

$\large 2x^2-3x+9=0$

By the quadratic equation formula , we get $x_1 = \dfrac {3 + i\sqrt{63}}4$ and $x_2 = \dfrac{3 - i \sqrt{63}}4$ .

$\begin{aligned} x_1^2 + x_2^2 & = \left(\dfrac{3 +i\sqrt{63}}4\right)^2 + \left( \dfrac{3 - i \sqrt{63}}4 \right)^2 \\ & = \dfrac{\left(3 + i\sqrt{63} \right)^2}{4^2} + \dfrac{(3- i\sqrt{63})^2}{4^2} \\ & = \dfrac{3^2 + 2 \cdot 3i\sqrt{63} +(i\sqrt{63})^2 }{16} + \dfrac{3^2 - 2 \cdot 3i\sqrt{63} + (i\sqrt{63})^2}{16} \\ & = \dfrac{9 + 6i\sqrt{63} - 63 + 9- 6i\sqrt{63} - 63}{16} &[i^2 = -1] \\ & = \dfrac{-108}{16} \\ & = \dfrac{-27}4. \\ \end{aligned}$