Suggested: 2x^2+x+4=0 - 2x-3/(x-2)(x-4) 0 - (2xy+x^2+x^4)dx-(1+x^2)dy=0 - 3/x-4/y-2/z-1=0 1/x+2/y+1/z-2=0 2/x-5/y-4/z+1=0 - 2x^2+x-4=0 complete the square - 2x^2+x-4=0 by factorisation method - x-y+z=4 2x+y-3z=0 x+y+z=2 - (3x^2y^4+2xy)dx+(2x^3y^3-x^2)dy=0 - solve the equation 2x^5+x^4-12x^3-12x^2+x+2=0 - if alpha beta are roots of the equation x^2-2x+4=0 - x+2y+2z=2 3x-2y-z=5 2x-5y+3z=-4 x+4y+6z=0 - the sum of the cubes of all the roots of the equation x^(4)-3x^(3)-2x^(2)+3x+1=0 is - 2x^2+x+4=0 Browse related:
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