k^2-12k-69=0

Solution for k^2-12k-69=0 equation:

k^2-12k-69=0

a = 1; b = -12; c = -69;
Δ = b2-4ac
Δ = -122-4·1·(-69)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{105}}{2*1}=\frac{12-2\sqrt{105}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{105}}{2*1}=\frac{12+2\sqrt{105}}{2}
$