How can I solve this system of non-linear equations?

$\{\begin{array}{lcr}a+c& =& 0\\ b+ac+d& =& 6\\ bc+ad& =& -5\\ bd& =& 6\end{array}$

The book I'm working from provides the answer $a=-1,b=1,c=1,d=6$, but I'm having a hard time getting there.

This is in order to reduce $q(x)={x}^{4}+6{x}^{2}-5x+6$ to a pair of irreducible quadratics $({x}^{2}+ax+b)({x}^{2}+cx+d)$. We expand, gather like terms, and equate the coefficients of the resulting expression with those of the original, yielding the above system of equations.

With some substitutions, I managed to get

${b}^{6}-6{b}^{5}-6{b}^{4}+47{b}^{3}-36{b}^{2}-216b+216=0$

which I reduced to

$(b-6)(b-1)({b}^{4}+{b}^{3}-5{b}^{2}+6b+36)=0$

Since the remaining factor $({b}^{4}+{b}^{3}-5{b}^{2}+6b+36)$ has no real zeros, it also reduces to a pair of irreducible quadratics.