k2-16k+23=0

Solution for k2-16k+23=0 equation:

k2-16k+23=0

We add all the numbers together, and all the variables

k^2-16k+23=0

a = 1; b = -16; c = +23;
Δ = b2-4ac
Δ = -162-4·1·23
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{41}}{2*1}=\frac{16-2\sqrt{41}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{41}}{2*1}=\frac{16+2\sqrt{41}}{2}
$