If it's not what You are looking for type in the equation solver your own equation and let us solve it.
x in (-oo:+oo)
a^3*((3*a^6*b^2*c^5)/9)*b^4*c^7 = 0
1/3*a^9*b^6*c^12 = 0
x należy do R
x in (-oo:+oo)
| x^3+3*x^2-4*x-12=0 | | X=5(y-z) | | 11y-6y-11=45.25 | | 1/2x1/3=N/24 | | x^2+5x-2=3x-2 | | 4x^2-2+3x=0 | | 20x-8=14x+21 | | 0.36/0.9=N/0.5 | | 16n+4=12 | | 9v^2+39v+12=0 | | 16=-2(2x-4) | | 0.24/0.8=n/6 | | 15n^2+57n+36=0 | | 8*3-6=-2y | | 8*2-6=-2y | | T=2s+3.63 | | 8*1-6=-2y | | 3x-5+3x=3(6x-4)+3 | | 8*3-y=16 | | 6-7x=5(1-8x) | | 8*2-y=16 | | 2x-4(x-4)=9+2x+1 | | 28x=4x+408 | | 8*1-y=16 | | 1/6=x-5 | | (3x+9)(x+1)=0 | | 5(x-6)+4=7x+10 | | 5x+5=5(x-3) | | 20y+4x=-32 | | 17/4x-2 | | x2/9=16 | | (x-5)(y-9)=0 |