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Use the Lagrange multiplier with the sign conventionFor every $\lambda >0$, this is a strictly convex quadratic in $x$, with unique global minimizerThese formulas follow by setting the three partial derivatives of the optimization Lagrangian to zero; its Hessian is $\mathrm{diag}(2\lambda ,2\lambda ,4)$, which is positive-definite.

Choose $\lambda >0$ to make this minimizer feasible. Its constraint value is $5/(2{\lambda }^2)-\lambda /4$, so the scalar equation isThe left side of the first equation is strictly decreasing on $(0,\infty )$, with limits $+\infty$ and $-\infty$. Thus exactly one positive multiplier exists and can be obtained by a bracketed scalar root search. The Lagrangian sufficiency theorem proves that its associated feasible point is the global optimum; strict Lagrangian convexity makes that optimum unique.

Consequently the unique solution isIts objective value is $-5/\lambda +{\lambda }^2/8$. This scalar multiplier certificate for a quadratic equality constraint does not require finding the multiplier in radicals, and does not mistake the nonconvex equality surface for a convex set.