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2k^2-13k+12=0
a = 2; b = -13; c = +12;
Δ = b2-4ac
Δ = -132-4·2·12
Δ = 73
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{73}}{2*2}=\frac{13-\sqrt{73}}{4} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{73}}{2*2}=\frac{13+\sqrt{73}}{4} $
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