Question

How to solve and why the answer is (A)

Answer 1

$\dfrac{1}{2}x^2+mx=5\\ \dfrac{1}{2}x^2+mx-5=0\\\\ \Delta=m^2-4\cdot\dfrac{1}{2}\cdot(-5)=m^2+10\\ \sqrt{\Delta}=\sqrt{m^2+10}\\\\ x=\dfrac{-m\pm\sqrt{m^2+10}}{2\cdot\dfrac{1}{2}}=-m\pm\sqrt{m^2+10}$