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-k+k(k+1)=2k
We move all terms to the left:
-k+k(k+1)-(2k)=0
We add all the numbers together, and all the variables
-3k+k(k+1)=0
We multiply parentheses
k^2-3k+k=0
We add all the numbers together, and all the variables
k^2-2k=0
a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
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{4}=2$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*1}=\frac{0}{2} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2} =2 $
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