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k^2+7k+6=-4
We move all terms to the left:
k^2+7k+6-(-4)=0
We add all the numbers together, and all the variables
k^2+7k+10=0
a = 1; b = 7; c = +10;
Δ = b2-4ac
Δ = 72-4·1·10
Δ = 9
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{9}=3$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3}{2*1}=\frac{-10}{2} =-5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3}{2*1}=\frac{-4}{2} =-2 $
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