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2k^2-5k-18=0
a = 2; b = -5; c = -18;
Δ = b2-4ac
Δ = -52-4·2·(-18)
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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*2}=\frac{-8}{4} =-2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*2}=\frac{18}{4} =4+1/2 $
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