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6k^2+11k+3=0
a = 6; b = 11; c = +3;
Δ = b2-4ac
Δ = 112-4·6·3
Δ = 49
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{49}=7$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*6}=\frac{-18}{12} =-1+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*6}=\frac{-4}{12} =-1/3 $
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