Solving for H+ given dissociation of weak acid, c and Ka Can the quadratic equation yield two nontrivial solutions?
What you have got here is a strong acid ($\mathrm{p}K_\mathrm{a} \approx 0.38$) that is not diluted, whereof the latter tells you that you can safely ignore the autoprotolysis of water.

The pH approximation to choose is therefor:
$$[\ce{H+}] = \frac{[\ce{HA}]_0 K_\mathrm{a}}{[\ce{H+}] + K_\mathrm{a}}$$

or in a quadratic form:
$$[\ce{H+}]^2 + [\ce{H+}] K_\mathrm{a} - [\ce{HA}]_0 K_\mathrm{a} = 0$$

which has two solutions:
$$[\ce{H+}] = -\frac{K_\mathrm{a}}{2} \pm \sqrt{\left(\frac{K_\mathrm{a}}{2}\right)^2 + [\ce{HA}]_0 K_\mathrm{a}}$$

From those two solutions, only one is chemically meaningful, which is the positive combination which yields one of your solutions ... $[\ce{H+}] = 0.0834$.