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2k^2-11k-6=0
a = 2; b = -11; c = -6;
Δ = b2-4ac
Δ = -112-4·2·(-6)
Δ = 169
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{169}=13$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*2}=\frac{-2}{4} =-1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*2}=\frac{24}{4} =6 $
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