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6k^2+3k-84=0
a = 6; b = 3; c = -84;
Δ = b2-4ac
Δ = 32-4·6·(-84)
Δ = 2025
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{2025}=45$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-45}{2*6}=\frac{-48}{12} =-4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+45}{2*6}=\frac{42}{12} =3+1/2 $
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