(2) \( 8 x=-12 x^{2} \) (5)* \( x^{3}=36 x \) (8)* \( 3 x^{2}-15=0 \) (2) \( x^{2}+12=7 x \) (5) \( 2 x^{2}-7 x=9 \) (8) \( 8 x^{2}+6 x=9 \) (11) \( 2 x-5 x^{2}+7=0 \)

Let’s tackle those quadratic equations! For \( 8 x=-12 x^{2} \), rearranging gives \( 12 x^{2} + 8 x = 0 \). Factoring out \( 4x \) leads to \( 4x(3x + 2) = 0 \), giving the solutions \( x = 0 \) or \( x = -\frac{2}{3} \). Now for \( x^{3}=36 x \), rewrite it as \( x^{3} - 36 x = 0 \) or \( x(x^{2}-36) = 0 \). So, \( x = 0 \) or \( x = 6, -6 \)! To solve \( 3 x^{2}-15=0 \), add 15 to both sides, yielding \( 3 x^{2} = 15 \). Divide by 3 gives \( x^{2} = 5 \), so \( x = \sqrt{5} \) or \( x = -\sqrt{5} \). Next, from \( x^{2}+12=7 x \), rearranging gives \( x^{2}-7x+12=0 \), which factors to \( (x-3)(x-4) = 0 \). The solutions are \( x = 3 \) or \( x = 4 \). For \( 2 x^{2}-7 x=9 \), rearranging leads to \( 2 x^{2} - 7 x - 9 = 0 \). Applying the quadratic formula \( x = \frac{-(-7) \pm \sqrt{(-7)^{2} - 4 \cdot 2 \cdot (-9)}}{2 \cdot 2} \) gives two solutions! In \( 8 x^{2}+6 x=9 \), shift 9 over yielding \( 8 x^{2} + 6x - 9 = 0 \) and again use the quadratic formula for solutions. Finally, for \( 2 x-5 x^{2}+7=0 \), rearranging yields \( -5 x^{2} + 2 x + 7 = 0 \). Applying the quadratic formula will help discover the roots of this quirky quadratic! Happy solving!