10k(k+2)=k-6

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Solution for 10k(k+2)=k-6 equation:



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 = b2-4ac
Δ = 192-4·10·6
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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