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3k^2-13k-10=0
a = 3; b = -13; c = -10;
Δ = b2-4ac
Δ = -132-4·3·(-10)
Δ = 289
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{289}=17$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*3}=\frac{-4}{6} =-2/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*3}=\frac{30}{6} =5 $
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