Question 628978
{{{x(2x^2+9x-56)(3x+10)=0}}}
I did this:
(2x^3+9x^2-56x)(3x+10)=0
6x^4+27x^3-168x^2+20x^3+90x^2-560x=0
6x^4+47x^3-78x^2-560x=0
6x^3+47x^2-78x-560=0

But I do not know what to do afterwards (I am supposed to solve for x).
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Go back to this:
{{{x(2x^2+9x-56)(3x+10)=0}}}
x = 0
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{{{(2x^2+9x-56)(3x+10)=0}}}
3x+10 = 0
x = -10/3
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{{{(2x^2+9x-56)=0}}}
Solve the remaining quadratic.