Witaj :)
[tex]\frac{1}{2}x^2-3x+9=(x-3)(x-5)\\\\\frac{1}{2}x^2-3x+9=x^2-5x-3x+15\\\\\frac{1}{2}x^2-x^2-3x+5x+3x+9-15=0\\\\-\frac{1}{2}x^2+5x-6=0\ / \cdot (-2)\\\\x^2-10x+12=0\\\\a=1,\ b=-10,\ c=12\\\\\Delta = b^2-4ac = (-10)^2-4\cdot 1\cdot 12=100-48=52 >0\\\\\sqrt{\Delta}=\sqrt{52} = \sqrt{4\cdot13} =2\sqrt{13}\\\\x_1= \frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{13} }{2}=\frac{10-2\sqrt{13} }{2}=5-\sqrt{13} \\\\[/tex]
[tex]x_2= \frac{-b+\sqrt{\Delta} }{2a}=\frac{-(-10)+2\sqrt{13} }{2} =\frac{10+2\sqrt{13} }{2}=5+\sqrt{13}[/tex]
ODP.:
[tex]\boxed {x \in \{5-\sqrt{13}\ ; 5+ \sqrt{13} \}}[/tex]
