Question 852178
In order to solve for x, the first thing you want to do is set this expression equal to 0:


{{{5x^3+30x^2-15x=0}}}


Next, since each of these terms have 5x in common, we can factor 5x out of the equation, giving us:


{{{5x(x^2+6x-3)=0}}}


Setting each of these terms equal to 0, gives us:


{{{5x=0}}} and {{{x^2+6x-3=0}}}


With 5x = 0, we can quickly see that x = 0, so, we know that one of the values of x is 0.


Now, in order to find the other two values of x, we need to use the quadratic formula on our second term, x^2 + 6x - 3 = 0:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} ----->


{{{x = (-6 +- sqrt( 6^2-4*1*-3 ))/(2*1) }}} ----->


{{{x = (-6 +- sqrt( 36+12 ))/(2) }}} ----->


{{{x = (-6 +- sqrt( 48 ))/(2) }}} ----->


{{{x = (-6 +- 4*sqrt( 3 ))/(2) }}}


Now, we can factor a 2 out of the numerator, since 2 is a factor of both -6 and +- 4:


{{{x = (2(-3 +- 2*sqrt( 3 )))/(2) }}} ----->


{{{x = -3 +- 2*sqrt( 3 ) }}}


We now have our 3 values of x: 0,{{{-3+2*sqrt(3)}}},{{{-3-2*sqrt(3)}}}