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3k^2+2k-208=0
a = 3; b = 2; c = -208;
Δ = b2-4ac
Δ = 22-4·3·(-208)
Δ = 2500
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{2500}=50$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-50}{2*3}=\frac{-52}{6} =-8+2/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+50}{2*3}=\frac{48}{6} =8 $
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