k^2+14k-22=6

Solution for k^2+14k-22=6 equation:

k^2+14k-22=6

We move all terms to the left:

k^2+14k-22-(6)=0

We add all the numbers together, and all the variables

k^2+14k-28=0

a = 1; b = 14; c = -28;
Δ = b2-4ac
Δ = 142-4·1·(-28)
Δ = 308
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:

$\sqrt{\Delta}=\sqrt{308}=\sqrt{4*77}=\sqrt{4}*\sqrt{77}=2\sqrt{77}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{77}}{2*1}=\frac{-14-2\sqrt{77}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{77}}{2*1}=\frac{-14+2\sqrt{77}}{2}
$