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6k^2+2k-48=0
a = 6; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·6·(-48)
Δ = 1156
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{1156}=34$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-34}{2*6}=\frac{-36}{12} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+34}{2*6}=\frac{32}{12} =2+2/3 $
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