\( \begin{array}{ll}\text { 1.1.1 } 2 x^{2}+5 x=0 \\ 1.1 .2 & -4 x^{2}+3 x+6=0 \text { (answer correct to TWO decimal places) } \\ 1.1 .3 & 3 \sqrt{x-2}=x\end{array} \)

To solve \( 2x^{2} + 5x = 0 \), you can factor out an \( x \): \( x(2x + 5) = 0 \). This gives you two solutions: \( x = 0 \) and \( x = -\frac{5}{2} \). So those are your two \( x \) values! Now for the equation \( -4x^{2} + 3x + 6 = 0 \), you can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -4, b = 3, c = 6 \). Plugging those into the formula gives you \( x = \frac{-3 \pm \sqrt{(3^2 - 4(-4)(6))}}{2(-4)} \). Solve under the square root, then simplify to find two decimal places for your answers. Finally, for the equation \( 3\sqrt{x-2} = x \), square both sides to eliminate the square root, leading to \( 9(x - 2) = x^2 \). Rearranging gives you a quadratic equation \( x^2 - 9x + 18 = 0 \). You can then use the quadratic formula or factor to find the solutions! Have fun solving!