If it's not what You are looking for type in the equation solver your own equation and let us solve it.
1*x^4-(6*x^3)+9*x^2-(4*x)+0 = 0;x
x
x^2+8;x^2+1
-((x^2+x+1)*(x^2-x+1)) < 0;x
x
5;8;13;32;0;0
-((x^2+x+1)*(x^2-x+1)) = 0;x
x
| -45=-9(2x-3) | | 8x-5(-4x+3)=13 | | 9a-3a+8= | | y=-2(-3)+2 | | 2x^2+4=y-15 | | 5z+3+2z+1= | | 6x^2+6-5x=0 | | 17u=9u+64 | | 9.2-1.7x=2.3x+1.2 | | (3x^2)+6x=45 | | -1/5b=-1/5 | | 2y+14=3y+2 | | 2+5=50 | | k/12=-9 | | 0.09xX=131.65 | | 3[2m-9+6m]=3m+29-7m | | 12a+10b+6.40c=300 | | 2n+3n-2n= | | .8(x+4000)=4000x+1000 | | 2[3m-6+3m]=2m+2-4m | | 0=5(x-7)+12 | | C/2-5=3 | | 66=5(x-7)+12 | | 10psquared+4p+77=9 | | 2/p=6/9 | | 2y-3+4w-2y= | | x/43-87=-755 | | 7y-3=22 | | 8a+2=6a-4 | | Z^4x^5/zx^7 | | 2y-x+1=0 | | a+6x=x+7 |