Since imaginary roots occur in the form of conjugate pairs then the other two roots will be
"1-\\sqrt3 \\ \\ \\ \\& \\ \\ \\ 2+i"
Sum of roots = "6"
products of roots = -10
"\\varSigma\\alpha\\beta=(1+\\sqrt3)\\{ 1-\\sqrt3+(2+i)+(2-i)\\}+(1-\\sqrt3)\\{(2+i)+(2-i)\\}+(2+i)(2-i)\\\\=2+4\\sqrt3+4-4\\sqrt3+5=11"
"\\varSigma\\alpha\\beta\\gamma=(1+\\sqrt3)(1-\\sqrt3)(2+i)+(1+\\sqrt3)(1-\\sqrt3)(2-i)+(1+\\sqrt3)(2+i)(2-i)+(1-\\sqrt3)(2+i)(2-i)\\\\=-4-2i-4+2i+5+5\\sqrt3+5-5\\sqrt3=2"
equation: "x^4-6x^3+11x^2-2x-10=0"
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