What are the real roots of 2x^3+9x^2+4x-15=0

Well, luckily it is apparent that (x-1) is a root because when x=1 the equation is equal to zero. So we can divide the equation by that factor to find the other roots.

(2x^3+9x^2+4x-15) / (x-1)

2x^2 r 11x^2+4x-15

11x r 15x-15

15 r 0

(x-1) (2x^2+11x+15) = 0

(x-1) (2x^2+6x+5x+15) = 0

(x-1) (2x (x+3) + 5 (x+3)) = 0

(x-1) (2x+5) (x+3) = 0

So the roots are x = - 3, - 2.5, 1