Question

$x^2+x+\frac{1}{\sqrt{2}}=0$

Answer 1

$x^2+x+\frac{1}{\sqrt{2}}=0\text{We will apply discriminant rule,}x=\frac{-b\pm \sqrt{D}}{2a}...\left(A\right)whereD=b^2-4ac=1^2-4.1\frac{1}{\sqrt{2}}=1-2\sqrt{2}=-7from\left(A\right)x=\frac{-1\pm \sqrt{-\left(2\sqrt{2-1}\right)}}{2}=\frac{-1\pm \sqrt{2\sqrt{2}-1}}{2}=\frac{-1\pm \sqrt{2\sqrt{2}-1i}}{2}Thus,\therefore x=\frac{-1\pm \sqrt{2}\sqrt{2}-1i}{2}$