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k^2=-42+13k
We move all terms to the left:
k^2-(-42+13k)=0
We add all the numbers together, and all the variables
k^2-(13k-42)=0
We get rid of parentheses
k^2-13k+42=0
a = 1; b = -13; c = +42;
Δ = b2-4ac
Δ = -132-4·1·42
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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-1}{2*1}=\frac{12}{2} =6 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*1}=\frac{14}{2} =7 $
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