[tex]x^2-10x+25=18 \\ \\x^2-10x+25-18 = 0\\ \\ x^2-10x+ 7 =0 \\ \\a=1 , b = -10 , c= 7 \\ \\ \Delta = b^{2}-4ac = (-10)^{2}-4*1*7=100-28 = 72 \\ \\\sqrt{\Delta }=\sqrt{72}= \sqrt{2*36} =\sqrt{36}*\sqrt{2} =6\sqrt{2}[/tex]
[tex]x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{10- 6\sqrt{2}}{2}=\frac{2(5-3\sqrt{2})}{2}= 5-3\sqrt{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{10+ 6\sqrt{2}}{2}=\frac{2(5+3\sqrt{2})}{2}= 5+3\sqrt{2}[/tex]
[tex]Answer : \ x= 5-3\sqrt{2} \ \ or \ \ x = 5+3\sqrt{2}[/tex]
