(k)^2=2-k

Solution for (k)^2=2-k equation:

(k)^2=2-k

We move all terms to the left:

(k)^2-(2-k)=0

We add all the numbers together, and all the variables

k^2-(-1k+2)=0

We get rid of parentheses

k^2+1k-2=0

We add all the numbers together, and all the variables

k^2+k-2=0

a = 1; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*1}=\frac{-4}{2}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*1}=\frac{2}{2}
=1
$