how do u slove 2x^2-3x-9
\(2x^2-3x-9 = 2x^2-6x + 3x -9 = 2x(x-3) + 3(x-3) = (2x+3)(x-3)\)
Alternatively, I could have used the quadratic formula. Anyway, the above equation dictates that the two solutions are \(-\frac{3}{2} \& 3\)
