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OpenStudy (anonymous):
Solve: x(3x+1)(2x-5)=0
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OpenStudy (tfht):
(3x^2+x)(2x-5)= (6x^3)+(2x^2)+(15x^2)+(-5x) (6x^3)+(17x^2)+(-5x)=0 x((6x^2)+(17x)-5 so then set everything to 0
OpenStudy (tfht):
wait nevermind... im lost
OpenStudy (tfht):
o set everything equal to 5 so x=5, x=the square root of 5/6 and x=5/17. I think
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