[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^2-3x+9=x^2-8x+15\ \ |*2\\\\x^2-6x+18=2x^2-16x+30\\\\x^2-6x+18-2x^2+16x-30=0\\\\-x^2+10x-12=0\ \ |*(-1)\\\\ x^2-10x+12=0[/tex]
[tex]a=1,\ \ b=-10,\ \ c=12 \\\\\Delta =b^2-4ac=(-10)^2-4*1*12=100-48=52\\\\\sqrt{\Delta }=\sqrt{52}=\sqrt{4*13}=2\sqrt{13}\\\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{10-2\sqrt{13}}{2*1}=\frac{ \not{2}^1 (5- \sqrt{13} )}{\not{2}^1 }=5-\sqrt{13}\\\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{10+2\sqrt{13}}{2*1}=\frac{ \not{2}^1 (5+ \sqrt{13} )}{\not{2}^1 }=5+\sqrt{13}[/tex]
