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k^2+6k+5=-3
We move all terms to the left:
k^2+6k+5-(-3)=0
We add all the numbers together, and all the variables
k^2+6k+8=0
a = 1; b = 6; c = +8;
Δ = b2-4ac
Δ = 62-4·1·8
Δ = 4
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{4}=2$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*1}=\frac{-8}{2} =-4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*1}=\frac{-4}{2} =-2 $
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