-q2+25q-46=0

Solution for -q2+25q-46=0 equation:

-q2+25q-46=0

We add all the numbers together, and all the variables

-1q^2+25q-46=0

a = -1; b = 25; c = -46;
Δ = b2-4ac
Δ = 252-4·(-1)·(-46)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-21}{2*-1}=\frac{-46}{-2}
=+23
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+21}{2*-1}=\frac{-4}{-2}
=+2
$