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6k^2+19k+10=0
a = 6; b = 19; c = +10;
Δ = b2-4ac
Δ = 192-4·6·10
Δ = 121
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{121}=11$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*6}=\frac{-30}{12} =-2+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*6}=\frac{-8}{12} =-2/3 $
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