Question 439844
{{{6x^3-150x=0}}}, first we factor out 6x: {{{6x(x^2-25)=0}}}, factoring further

{{{6x(x-5)(x+5)=0}}}. Now we set to zero three factors of this product:

6x=0=> x=0; x-5=0 => x=5; x+5=0 => x=-5.

Answer:The roots of this equation are: {-5, 0, 5}.