# Revision history [back]

I'm not sure I completely understand your question, but perhaps this is _a_ way of solving for the answer you seek:

[myroot.n() for myroot,mymult in (x^3 + 4*x - 7).roots() if myroot.n().parent() == RR]


I'm not sure I completely understand your question, but perhaps this is _a_ way of solving for the answer you seek:

[myroot.n() for myroot,mymult in (x^3 + 4*x - 7).roots() if myroot.n().parent() == RR]
[1.25538315684475]


I'm not sure I completely understand your question, but perhaps this is _a_ way of solving for the answer answers you seek:

[myroot.n() for myroot,mymult in (x^3 + 4*x - 7).roots() if myroot.n().parent() == RR]
[1.25538315684475]

[mysol.rhs().n() for mysol in solve(x^3 + 4*x - 2 == -5, x) if mysol.rhs().n().parent() == RR]
[?0.673593058218711]


I'm not sure I completely understand your question, but perhaps this is _a_ way of solving for the answers you seek:

[myroot.n() for myroot,mymult in (x^3 + 4*x - 7).roots() if myroot.n().parent() == RR]
[1.25538315684475]

[mysol.rhs().n() for mysol in solve(x^3 + 4*x - 2 == -5, x) if mysol.rhs().n().parent() == RR]
[?0.673593058218711]
[0.673593058218711]


I'm not sure I completely understand your question, but perhaps this is _a_ way of solving for the answers you seek:

[myroot.n() for myroot,mymult in (x^3 + 4*x - 7).roots() if myroot.n().parent() == RR]
[1.25538315684475]

[mysol.rhs().n() for mysol in solve(x^3 + 4*x - 2 == -5, x) if mysol.rhs().n().parent() == RR]
[0.673593058218711]
[-0.673593058218711]