Let's solve in $ℝ$: $-x^2+x-1<0$

Given the quadratic form $a{x}^2+bx+c$, you can calculate the discriminant: $\Delta =b^2-4ac$. Here $a=-1$, $b=1$ and $c=-1$

$\Delta =1-4·\left(-1\right)·\left(-1\right)$

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive: $-4·\left(-1\right)$ = $4$■Multiplying by -1 gives the opposite: $4·\left(-1\right)$ = $-4$

$\Delta =1-4$

ⓘ Start from $1$ and move back $4$ on the number line, going below zero: $1-4$ = $-3$

$\Delta =-3$

The quadratic has no real root: it has the sign of a = $-1$ everywhere, so it is indeed negative for every $x$.

▶Solution found: $x\in ℝ$