If it's not what You are looking for type in the equation solver your own equation and let us solve it.
1;3;0;0;-4;3
(1/4)*(x-2)-((2/5)*(x+2)) = (1/10)*(x-1)+1/20;x
x
sin(x^2)
1;3;0;0;0;1
x^3+4*x^2+x = 0;x
x
| a-12=-9 | | (0.5x^4)+(x^3)+1=(0.5x^4)+(x^3)+(x^2)-mx+1 | | Cos(3x)=1/rad2 | | 7x-5+2x=3x+10 | | 7k+0=7k | | x+(x*.25)=145 | | (5g+3)=(g+8) | | 7(18-x)+2x=39 | | 1=x^2-mx+2 | | -7.6=x-1.4 | | 7x-6=10x+16 | | 3(4t+1)=78 | | -4.1x+3.6=-21 | | 3y=25-2y | | 5/1x1/2 | | x^11/x^9 | | 20x+3=51 | | 11+x=15x | | 5x-2=7x-9 | | 21=x-[-4] | | 700-35x=450 | | 2x+4y(91-x)=280 | | (1+u)(5u+9)=0 | | 0.666-10x=2.333 | | 2x-4x-16=0 | | n+[-20]=-32 | | -2x+0.75=0.30 | | n-[-16]=28 | | -7x+3=-17.50 | | 5-(j/12)=1 | | 8p-6=4p+3+4p | | 0.3=2.1 |