k2+24k+23=0

Solution for k2+24k+23=0 equation:

k2+24k+23=0

We add all the numbers together, and all the variables

k^2+24k+23=0

a = 1; b = 24; c = +23;
Δ = b2-4ac
Δ = 242-4·1·23
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-22}{2*1}=\frac{-46}{2}
=-23
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+22}{2*1}=\frac{-2}{2}
=-1
$