\( \begin{array}{ll}1.1 .1 & 2 x^{2}+5 x=0 \\ 1.1 .2 & -4 . x^{2}+3 x+6= \\ 1.1 .3 & 3 \sqrt{x-2}=x \\ \text { Given: } x^{2}-x-20<0 \\ 1.2 .1 \quad \text { Solve for } x \text { ir } x^{2}\end{array} \)

Let’s dive into these quadratic equations! The first equation, \(2x^{2} + 5x = 0\), can be factored as \(x(2x + 5) = 0\), leading to \(x = 0\) and \(x = -\frac{5}{2}\). This illustrates the knack of factoring to solve quadratics quickly! Next up, for \(3\sqrt{x-2} = x\), squaring both sides results in \(9(x - 2) = x^2\). Rearranging yields \(x^2 - 9x + 18 = 0\), which factors to \((x - 6)(x - 3) = 0\). Thus, \(x = 3\) or \(x = 6\) are your solutions. Always remember to check for extraneous solutions when squaring both sides!