-q^2+440q+42=0

Solution for -q^2+440q+42=0 equation:

-q^2+440q+42=0

We add all the numbers together, and all the variables

-1q^2+440q+42=0

a = -1; b = 440; c = +42;
Δ = b2-4ac
Δ = 4402-4·(-1)·42
Δ = 193768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{193768}=\sqrt{4*48442}=\sqrt{4}*\sqrt{48442}=2\sqrt{48442}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(440)-2\sqrt{48442}}{2*-1}=\frac{-440-2\sqrt{48442}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(440)+2\sqrt{48442}}{2*-1}=\frac{-440+2\sqrt{48442}}{-2}
$