a)
[tex]5x-x^{2} = 0 \ \ /\cdot(-1)\\\\x^{2}-5x = 0\\\\x(x-5) = 0\\\\x = 0 \ \vee \ x - 5 = 0\\\\x = 0 \ \vee \ x = 5\\\\x \in \{0,5\}[/tex]
b)
[tex]-x^{2}+5x+1 = 0 \ \ /\cdot(-1)\\\\\underline{x^{2}-5x-1 = 0}\\\\a =1, \ b = -5, \ c = -1\\\\\Delta = b^{2}-4ac = (-5)^{2}-4\cdot1\cdot(-1) = 25+4 = 29\\\\\sqrt{\Delta} = \sqrt{29}\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-5)-\sqrt{29}}{2\cdot1} = \frac{5-\sqrt{29}}{2}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{5+\sqrt{29}}{2}\\\\x \in \{\frac{5-\sqrt{29}}{2}, \frac{5+\sqrt{29}}{2}\}[/tex]
c)
[tex]2x^{2}-3x+3 = 0\\\\a = 2, \ b = -3, \ c = 3\\\\\Delta = b^{2}-4ac=(-3)^{2}-4\cdot2\cdot3 = 9 - 24 = -15 < 0[/tex]
Δ < 0, brak rozwiązań.
