Answer:
[tex]x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}[/tex]
Step-by-step explanation:
We need to solve the quadratic equation
6 = x^2 -10x
Rearranging we get,
x^2-10x-6=0
Using quadratic formula to solve the quadratic equation
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
a= 1, b =-10 and c=6
Putting values in the quadratic formula
[tex]x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}[/tex]
So, [tex]x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}[/tex]
