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6k^2-17k+7=0
a = 6; b = -17; c = +7;
Δ = b2-4ac
Δ = -172-4·6·7
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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-11}{2*6}=\frac{6}{12} =1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+11}{2*6}=\frac{28}{12} =2+1/3 $
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