k^2+8k-23=0

Solution for k^2+8k-23=0 equation:

k^2+8k-23=0

a = 1; b = 8; c = -23;
Δ = b2-4ac
Δ = 82-4·1·(-23)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{39}}{2*1}=\frac{-8-2\sqrt{39}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{39}}{2*1}=\frac{-8+2\sqrt{39}}{2}
$