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10k(k+2)=k-6

We move all terms to the left:

10k(k+2)-(k-6)=0

We multiply parentheses

10k^2+20k-(k-6)=0

We get rid of parentheses

10k^2+20k-k+6=0

We add all the numbers together, and all the variables

10k^2+19k+6=0

a = 10; b = 19; c = +6;

Δ = b^{2}-4ac

Δ = 19^{2}-4·10·6

Δ = 121

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{121}=11$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*10}=\frac{-30}{20} =-1+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*10}=\frac{-8}{20} =-2/5 $

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