k^2+2k-13=0

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

k^2+2k-13=0

a = 1; b = 2; c = -13;
Δ = b2-4ac
Δ = 22-4·1·(-13)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{14}}{2*1}=\frac{-2-2\sqrt{14}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{14}}{2*1}=\frac{-2+2\sqrt{14}}{2}
$