Bonsoir,
[tex]3x^2-16x+16=0\\\\
\Delta = b^2-4ac\\
\Delta = (-16)^2-4\times3\times16\\
\Delta = 256 -192\\
\Delta = 64\\\\
x_1 = \dfrac{-b-\sqrt{\Delta}}{2a} \quad \quad x_2 = \dfrac{-b+\sqrt{\Delta}}{2a}\\\\
x_1 = \dfrac{-(-16)-\sqrt{64}}{2\times3} \quad \quad x_2 = \dfrac{-(-16)+\sqrt{64}}{2\times3}\\\\
x_1 = \dfrac{16-8}{6} \quad \quad x_2 = \dfrac{16+8}{6}\\\\
x_1 = \dfrac{8}{6} \quad \quad x_2 = \dfrac{24}{6}\\\\
x_1 = \dfrac{4}{3} \quad \quad x_2 = 4\\\\
\boxed{S = \left\{ \dfrac{4}{3}; 4\right\}}[/tex]
