Question

$\sqrt{5}{x}^2+x+\sqrt{5}=0$

Answer 1

$\sqrt{5}{x}^2+x+\sqrt{5}=0\text{We will apply discriminant rule,}x=\frac{-b\pm \sqrt{D}}{2a}...\left(A\right)whereD=b^2-4ac=1^2-4.\sqrt{5}.\sqrt{5}=1-20=-19from\left(A\right)x=\frac{-1\pm \sqrt{-19}}{2.\sqrt{5}}=\frac{-1\pm \sqrt{19i}}{2\sqrt{5}}Thus,\therefore x=\frac{-1\pm \sqrt{19i}}{2\sqrt{5}}$