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k^2+2=-3k
We move all terms to the left:
k^2+2-(-3k)=0
We get rid of parentheses
k^2+3k+2=0
a = 1; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·1·2
Δ = 1
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{1}=1$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*1}=\frac{-4}{2} =-2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*1}=\frac{-2}{2} =-1 $
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