Question

14 >= 2x^2+8x-10

Here's as far as I got:

0>=2x^2+8x-24
0>=(2x-4)(x+6)

How do I get the values of x?

Answer 1

Find the values of x which make the equation equal to zero. That's easy, you set 2x-4 = 0, and x+6=0 and get the two values of x which solve those. So those are the two points where the function crosses over the x axis. From there, you could see that it's a quadratic with positive leading coefficient (the number in front of x^2 is positive) and know that those functions are are positive for large negative values of x, and positive for large positive values of x, or, you could choose three points, to the left, in the middle, and to the right of the two values of x where the function cross the x axis, and see if it's positive of negative in that area.... You want the points of x which make it less than or equal to zero... so.... they will be the points in between the two critical values of x...