k2+12k-14=0

Solution for k2+12k-14=0 equation:

k2+12k-14=0

We add all the numbers together, and all the variables

k^2+12k-14=0

a = 1; b = 12; c = -14;
Δ = b2-4ac
Δ = 122-4·1·(-14)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-10\sqrt{2}}{2*1}=\frac{-12-10\sqrt{2}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+10\sqrt{2}}{2*1}=\frac{-12+10\sqrt{2}}{2}
$