\( 1 \leftarrow \quad \begin{array}{l}\text { Solve } \\ 10 x^{4}-19 x^{2}+6=0\end{array} \)

To solve the equation \(10x^4 - 19x^2 + 6 = 0\), we can make a substitution. Let \(y = x^2\), transforming the equation into a quadratic form: \(10y^2 - 19y + 6 = 0\). Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we substitute \(a = 10, b = -19, c = 6\) to find the roots for \(y\). With the discriminant calculated as \((-19)^2 - 4 \cdot 10 \cdot 6 = 361 - 240 = 121\), we proceed with finding \(y\): \[ y = \frac{19 \pm \sqrt{121}}{20} = \frac{19 \pm 11}{20} \] This gives us two possible values for \(y\): \(y_1 = \frac{30}{20} = 1.5\) and \(y_2 = \frac{8}{20} = 0.4\). Now, translating back to \(x\) gives us \(x^2 = 1.5\) (so \(x = \pm \sqrt{1.5}\)) and \(x^2 = 0.4\) (so \(x = \pm \sqrt{0.4}\)). Thus, the solutions are \(x = \pm\sqrt{1.5}, \pm\sqrt{0.4}\). And there you have it, your solutions flaunting like stars in the mathematical sky!