jmc

We subtract $\sqrt[3]{x}$ from both sides, giving $$\sqrt{x+12}=-\sqrt[3]{x}.$$Now, to remove radicals, we raise both sides to the sixth power, giving $$(x+12{)}^3={\left(\sqrt{x+12}\right)}^6={\left(-\sqrt[3]{x}\right)}^6=x^2.$$Expanding the left-hand side and subtracting $x^2$ will create a nasty cubic in $x$, so we first make the substitution $y=x+12,$ which turns our equation into $$y^3=(y-12{)}^2,$$or $$y^3-y^2+24y-144=0.$$To find the roots of this equation, note that for $y=0,$ the left-hand side is $-144,$ which is negative, while for $y=5,$ the left-hand side is $76,$ which is positive; therefore, there must be a root in the interval $(0,5).$ Trying integer roots in this interval, we find that $y=4$ is a root of the equation. Factoring out $y-4$ from the equation gives $$(y-4)(y^2+3y+36)=0.$$The discriminant of the quadratic $y^2+3y+36$ is $3^2-4\cdot 36=-135,$ which is negative, so the only real root of the equation is $y=4.$ Thus, $x=y-12=\boxed{-8},$ which can be checked to satisfy the original equation.