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k^2+6k+4=0
a = 1; b = 6; c = +4;
Δ = b2-4ac
Δ = 62-4·1·4
Δ = 20
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{5}}{2*1}=\frac{-6-2\sqrt{5}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{5}}{2*1}=\frac{-6+2\sqrt{5}}{2} $
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