If it's not what You are looking for type in the equation solver your own equation and let us solve it.
D( x )
3*x^3-(2*x^2)-(5*x) = 0
3*x^3-(2*x^2)-(5*x) = 0
3*x^3-(2*x^2)-(5*x) = 0
3*x^3-2*x^2-5*x = 0
3*x^3-2*x^2-5*x = 0
x*(3*x^2-2*x-5) = 0
3*x^2-2*x-5 = 0
DELTA = (-2)^2-(-5*3*4)
DELTA = 64
DELTA > 0
x = (64^(1/2)+2)/(2*3) or x = (2-64^(1/2))/(2*3)
x = 5/3 or x = -1
x = 0
x = 0
x in (-oo:-1) U (-1:0) U (0:5/3) U (5/3:+oo)
(6*x^3-(10*x^2))/(3*x^3-(2*x^2)-(5*x)) = 0
(6*x^3-10*x^2)/(3*x^3-2*x^2-5*x) = 0
6*x^3-10*x^2 = 0
2*x^2*(3*x-5) = 0
3*x-5 = 0 // + 5
3*x = 5 // : 3
x = 5/3
2*x^2*(x-5/3) = 0
3*x^3-2*x^2-5*x = 0
x*(3*x^2-2*x-5) = 0
3*x^2-2*x-5 = 0
DELTA = (-2)^2-(-5*3*4)
DELTA = 64
DELTA > 0
x = (64^(1/2)+2)/(2*3) or x = (2-64^(1/2))/(2*3)
x = 5/3 or x = -1
x*(x+1)*(x-5/3) = 0
(2*x^2*(x-5/3))/(x*(x+1)*(x-5/3)) = 0
( 2*x^2 )
2*x^2 = 0 // : 2
x^2 = 0
x = 0
( x-5/3 )
x-5/3 = 0 // + 5/3
x = 5/3
x in { 0}
x in { 5/3}
x belongs to the empty set
| 7y-4=35 | | 7y-4=-6y | | 7-(3x+2)2=5-x | | F(-1)=x^2-4x^2+X | | 13y/72-5/8 | | 9/20-9/10 | | F(3)=-X^2+4x+1 | | 8z^4(3z^2-7)= | | a^2-17a-12=0 | | a^2-17a=12 | | (2z-7)-3(3z+8)=z-9 | | 9y^2-22y+13=0 | | 5(x-3)/7=-2x | | (2z-7)-3(3z+8)=7-9 | | -4v(v-2)=-4v(v-8)-8v | | 25p-3=0.2 | | 32=x^2+2x-3 | | 56+4n=4n-8(3n+5) | | ln(x)=-3x | | Y^2-20y+25=0 | | 6x-x=28 | | 2(3x+5)=58+x | | 18-3=16x-4+12 | | 3x+2(3)=12 | | 5x+6y-y=20 | | 49x^3+196x^2-16x-64=0 | | 11(C+7)=220 | | 37=5+(n-1)2 | | 5c-3=3c+1 | | 2x-3=3(5x-12/10 | | 3/20x=9 | | 3a+7=a+15 |