If it's not what You are looking for type in the equation solver your own equation and let us solve it.
y = ((x^2-(3*x)+2)/x+1)^(1/2)
y = ((3*x+1)^1)/2
y = ((x^2-(3*x)+2)/x+1)^(1/2)
((p^-5*y)/12)*s^-9 = 0
2*((3/4)*x-1) = 0
| 2y-8x-5x+6=0 | | 66=6.6w | | 66=6.6 | | 4y^2+8y-21= | | 16x+10y=-3 | | p^-5y/12s^-9 | | 10x+1=3x-9 | | 0=r(r)-7.29 | | 14y^2+43y-21= | | 2x-9y+45=0 | | 0=r(r)-24.01 | | 2(3/4x-1) | | (5u+8)(2-u)=0 | | 2y^5+20y^4+50y^3= | | 4/3(x-2)-1/2=2(3/4x-1) | | log(x)=4/5 | | X^3-88= | | 2y^4+12y^3+18y^2= | | 149=x+(x*.825) | | x^2+y^2+10x+30y+100=0 | | 0=r(r)-19.36 | | 14-8=22+20p | | 3x/15*-5/2x*5x/30 | | y-4=1/2-1.5 | | 2(k-1.7)=-7.8(k-1.4) | | Sin^2x-4sinx-5=0 | | 8z-10=z-72+3 | | 5+a=16 | | y=(lne^x)/e^x | | x^5=174 | | 7.2a-0.7b-2.2+8.3b= | | x*.15=42 |