Example on how to solve polynomial

Introduction:

          In math, variables and constants with the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents of finite length is known as a polynomial or in-determinates. For example, x² − 4x + 7 is a polynomial, but x² − 4/x + 7x^3/2 is not, because its second term involves division by the variable x and because its third term contains an exponent that is not a whole number.

Example:

X²+4X-5=0,

                   X²-X+5X-5=0,

                   X(X-1) +5(X-1) =0,

                   (X-1)(X+5) =0.                         Therefore the roots are X =1, -5...

                       (OR)

USING THE FOMULAE:

                   X = ` (-B + sqrt (B^2 - 4AC))/(2A)`

                   X²+4X-5=0 -------------------> (2)

                    AX²+BX+C=0----------------> (3)

Comparing (2) and (3),

A = 1, B =4, C =-5 Substitute these values in one.

        x = `(-4 +- sqrt (4^2 - 4(1)(-5)))/(2(1))`

          x = `(-4 +- sqrt (16 +20))/(2)`

          x = `(-4 +- sqrt (36))/(2)`

         x = `(-4 +- 6)/(2)`

        x = `(-10)/(2)``(2)/(2)`

       x = -5, 1

Hope you like the above example. Please leave your comments, if you have any doubts.