-k^2+13k-42=0

Solution for -k^2+13k-42=0 equation:

-k^2+13k-42=0

We add all the numbers together, and all the variables

-1k^2+13k-42=0

a = -1; b = 13; c = -42;
Δ = b2-4ac
Δ = 132-4·(-1)·(-42)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*-1}=\frac{-14}{-2}
=+7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*-1}=\frac{-12}{-2}
=+6
$