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2k^2+13k-14=0
a = 2; b = 13; c = -14;
Δ = b2-4ac
Δ = 132-4·2·(-14)
Δ = 281
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{281}}{2*2}=\frac{-13-\sqrt{281}}{4} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{281}}{2*2}=\frac{-13+\sqrt{281}}{4} $
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