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)
(2*y+(3/4)*z)^2 = 0
(2*y+3/4*z)^2 = 0
x belongs to the empty set
| 221/n=13/12 | | n/3=300/12 | | 80/20 | | 4(2x-7)-8(5-x)=3(2x+4)-5(x+7) | | 52*7=3x | | Y=4x^2-6x-4 | | 3/8*m=-24 | | 3Y^(2/3)=192 | | 7/4=n/28 | | (-3x/x+12)=(9/x-18) | | 2n-5*-6=-20 | | 6(2+5x)= | | 5*2n-6=-20 | | 5/n=25/50 | | 9(x+2)=3(x+20) | | 3(x-3)=47-4x | | 8.6x-2.4x=3.6x | | 19/9x^10/9+10/9x^1/9 | | 2x+4(x-3)=0 | | n/40=10/20 | | 7-2(3-4x)= | | 12x=4x-8 | | 4x^2-18x-220=82 | | 8x-3x(4+x)= | | 2+32y= | | 4(3x-2)=6(x-3) | | 5/20=n/10 | | 8.09(A^11)-9.09(A^10)=-1 | | 5x+23=8x | | 24/5+13-3+14(4)=354/5 | | 6x^2-19xy-20y^2= | | .55x+.05(12-x)=.10(-34) |