Given the quadratic equation :

x^2 - 2x - 99 = 0

Let us factor.

==> (x -11)(x+ 9) = 0

Now we will find the roots.

==> x1= 11

==> x2= -9

Then the quadratic equation has two real root.

Then :

**(x^2 - 2x - 99) = ( x-11)(x+9) = 0 such that the roots are : -9 and 11.**

The roots of the equation (x^2 -2x - 99 ) = 0 have to be determined by factorization of the polynomial x^2 - 2x - 99.

To factor a polynomial of the form ax^2 + bx + c, write b as a sum of two terms t1 and t2 such that t1 + t2 = b and t1*t2 = a*c

x^2 - 2x - 99 = 0

We can write -2 as the sum of -11 and 9, notice that the product of -11 and 9 is -99*1

x^2 - 11x + 9x - 99 = 0

x(x - 11) + 9(x - 11) = 0

(x + 9)(x - 11) = 0

x + 9 = 0, x = -9

x - 11 = 0, x = 11

The solution of the equation x^2 - 2x - 99 = 0 = 0 is x = -9 and x = 11

We'll find out the roots of the equation, using quadratic formula:

x1 = [2 + sqrt(4+396)]/2

x1 = (2+20)/2

x1 = 11

x2 = (2-20)/2

x2 = -9

The quadratic can be written as a product of linear factors:

x^2 -2x - 99 = (x-x1)(x-x2)

x^2 -2x - 99 = (x-11)(x+9)

We'll cancel out the product:

(x-11)(x+9) = 0

x-11 = 0 => x1 = 11

x+9=0

x2 = -9

**The roots of the equation are: x1 = 11 and x2 = -9**

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