F(x)=x^2-2x-3

plz help me slove this ?

there's nothing to solve. You have just presented a function of x.

Now, if you want to find where f(x)=0, that's a problem you can solve. On the way, we'll discover how to factor the quadratic, which is another kind of problem.

F(x) = (x-3)(x+1)

so, F(x)=0 when

x-3 = 0
or
x+1 = 0

Those I think you can handle, eh?

OH OK THANK YOU

OH OK THANK YOU STEVE

To solve the equation f(x) = x^2 - 2x - 3, we need to find the values of x that satisfy this equation.

Step 1: Set the equation equal to zero.
x^2 - 2x - 3 = 0

Step 2: Factorize the quadratic expression if possible.
In this case, the expression cannot be easily factored. Therefore, we move to the next step.

Step 3: Use the quadratic formula.
The quadratic formula states:

For a quadratic equation in the form ax^2 + bx + c = 0, the solutions (or roots) can be found using the formula:

x = (-b ± √(b^2 - 4ac))/(2a)

For the equation x^2 - 2x - 3 = 0, a = 1, b = -2, c = -3.

x = (-(-2) ± √((-2)^2 - 4 *1 * -3))/(2*1)
x = (2 ± √(4 + 12))/2
x = (2 ± √(16))/2
x = (2 ± 4)/2

This gives us two potential solutions: x = (2 + 4)/2 and x = (2 - 4)/2.

Simplifying further:
x = 6/2 and x = -2/2

This leads to the solutions:
x = 3 and x = -1.

So, the solutions to the equation f(x) = x^2 - 2x - 3 are x = 3 and x = -1.