a)
[tex]81x^{2}-36 = 0 \ \ |:9\\\\9x^{2}-4 = 0\\\\(3x+2)(3x-2) = 0\\\\3x+2 = 0 \ \ \vee \ \ 3x-2 = 0\\\\3x = -2 \ \ |:3 \ \ \vee \ \ 3x = 2 \ \ |:3\\\\x = -\frac{2}{3} \ \ \vee \ \ x = \frac{2}{3}\\\\x\in\{-\frac{2}{3}, \frac{2}{3}\}[/tex]
b)
[tex]-x^{2}-\frac{3}{2}x + 10 = 0 \ \ |\cdot(-2)\\\\\underline{2x^{2}+3x-20 = 0}\\\\a = 2, \ b = 3, \ c = -20\\\\\Delta = b^{2}-4ac = 3^{2}-4\cdot2\cdot(-20) = 9 + 160 =169\\\\\sqrt{\Delta} = \sqrt{169} = 13\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-3-13}{2\cdot2} = \frac{-16}{4} = -4\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-3+13}{4} = \frac{10}{4}=\frac{5}{2} = 2,5\\\\x \in\{-4, \ 2,5\}[/tex]
c)
[tex]4x^{2}+2x-2 > 0 \ \ |:2\\\\\underline{2x^{2}+x-1 > 0}\\\\M. \ zerowe\\\\\Delta = b^{2}-4ac = 1^{2}-4\cdot2\cdot(-1) = 1+8 = 9\\\\\sqrt{\Delta} = \sqrt{9} = 3\\\\x_1 = \frac{-1-3}{4} = \frac{-4}{4} = -1\\\\x_2 = \frac{-1+3}{4} = \frac{2}{4} = \frac{1}{2}\\\\a > 0, \ zatem\\\\x\in(-\infty;-1) \ \cup \ (\frac{1}{2};+\infty)[/tex]