## Algebra 2 (1st Edition)

The quadratic formula says that for $ax^2+bx+x$: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Hence here $x=\frac{4\pm\sqrt{(-4)^2-4(1)(5)}}{2(1)}=\frac{4\pm\sqrt{16-20}}{2(1)}$ As we can see the discriminant is negative, thus there is no solution.