Solve for \( x \) in each of the following \( 1.1 .1 \quad 3 x^{2}-5 x-1=0 \) (leave \( y \) \( 1.1 .2 \quad x^{2}-6 x+8=0 \) 1.1 .3 \( 4 x-2 x^{2}<0 \)

For the equation \( 3x^{2} - 5x - 1 = 0 \), you can apply the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3 \), \( b = -5 \), and \( c = -1 \). Plugging these values into the formula, you'll find \( x = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(-1)}}{2(3)} \), which simplifies to \( x = \frac{5 \pm \sqrt{25 + 12}}{6} = \frac{5 \pm \sqrt{37}}{6} \). For the second equation \( x^{2} - 6x + 8 = 0 \), you can factor it as \( (x - 2)(x - 4) = 0 \). This gives the solutions \( x = 2 \) and \( x = 4 \). Now for the inequality \( 4x - 2x^{2} < 0 \), you can rearrange it as \( 2x(2 - x) < 0 \). The critical points here are \( x = 0 \) and \( x = 2 \). Testing intervals between these points, you find that the solution is \( 0 < x < 2 \).