Answer:
[tex]5 \pm 2\sqrt{3}[/tex]
Step-by-step explanation:
A quadratic equation in the form
[tex]ax^2+bx+c=0[/tex]
Can be solved by using the formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
In this problem, the equation is
[tex]x^2-10x+13=0[/tex]
So, we have
a = 1
b = -10
c = +13
Using the equation above, we find:
[tex]x=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(13)}}{2(1)}=\frac{10 \pm \sqrt{48}}{2}=\frac{10 \pm 4\sqrt{3}}{2}= 5 \pm 2\sqrt{3}[/tex]
