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k+2/5k=12
We move all terms to the left:
k+2/5k-(12)=0
Domain of the equation: 5k!=0We multiply all the terms by the denominator
k!=0/5
k!=0
k∈R
k*5k-12*5k+2=0
Wy multiply elements
5k^2-60k+2=0
a = 5; b = -60; c = +2;
Δ = b2-4ac
Δ = -602-4·5·2
Δ = 3560
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3560}=\sqrt{4*890}=\sqrt{4}*\sqrt{890}=2\sqrt{890}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-2\sqrt{890}}{2*5}=\frac{60-2\sqrt{890}}{10} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+2\sqrt{890}}{2*5}=\frac{60+2\sqrt{890}}{10} $
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