Let \( y \) be the new root. Then \( y = \frac{x+2}{x-1} \).\n\nSolving for \( x \) in terms of \( y \):\n\[ y(x-1) = x+2 \]\n\[ xy - y = x + 2 \]\n\[ x(y-1) = y+2 \]\n\[ x = \frac{y+2}{y-1} \]\n\nSubstitute this \( x \) back into the original quadratic equation \( ax^2+bx+c=0 \):\n\[ a\left(\frac{y+2}{y-1}\right)^2 + b\left(\frac{y+2}{y-1}\right) + c = 0 \]\n\[ a(y+2)^2 + b(y+2)(y-1) + c(y-1)^2 = 0 \]\n\nExpanding the terms:\n\[ a(y^2+4y+4) + b(y^2+y-2) + c(y^2-2y+1) = 0 \]\n\nGrouping by powers of \( y \):\n\[ y^2(a+b+c) + y(4a+b-2c) + (4a-2b+c) = 0 \]\n\nReplacing \( y \) with \( x \) gives the required equation:\n\[ (a+b+c)x^2 + (4a+b-2c)x + (4a-2b+c) = 0 \]