Find the number of real number solutions for the equation. x2 – 10x + 25 = 021those are my f answqer chgoices

Answer:Option B is correctThe number of real number solutions for the given equation is, 1.Step-by-step explanation:Given the equation: [tex]x^2-10x+25 =0[/tex]Since, this is a quadratic equation of the form of [tex]ax^2+bx+c =0[/tex]where a =1 , b = -10 and c =25.The discriminant of a quadratic equation is, [tex]{b^2-4ac}[/tex]then;Discriminant = [tex]{(-10)^2-4(1)(25)}[/tex] = (100-100) = 0Since, a  discriminant of zero means there is only one real solution for x.[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Substitute the given values.[tex]x = \frac{-(-10) \pm 0}{2(1)}[/tex][tex]x = \frac{10}{2}[/tex]Simplify:x = 5Therefore, the number of real solution for the given equation is, 1