k2+12k+23=0

Solution for k2+12k+23=0 equation:

k2+12k+23=0

We add all the numbers together, and all the variables

k^2+12k+23=0

a = 1; b = 12; c = +23;
Δ = b2-4ac
Δ = 122-4·1·23
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{13}}{2*1}=\frac{-12-2\sqrt{13}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{13}}{2*1}=\frac{-12+2\sqrt{13}}{2}
$