-q^2+q+4=0

Solution for -q^2+q+4=0 equation:

-q^2+q+4=0

We add all the numbers together, and all the variables

-1q^2+q+4=0

a = -1; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·(-1)·4
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{17}}{2*-1}=\frac{-1-\sqrt{17}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{17}}{2*-1}=\frac{-1+\sqrt{17}}{-2}
$