k2+2k-2=7

Solution for k2+2k-2=7 equation:

k2+2k-2=7

We move all terms to the left:

k2+2k-2-(7)=0

We add all the numbers together, and all the variables

k^2+2k-9=0

a = 1; b = 2; c = -9;
Δ = b2-4ac
Δ = 22-4·1·(-9)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*1}=\frac{-2-2\sqrt{10}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*1}=\frac{-2+2\sqrt{10}}{2}
$