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2k^2+3k-14=0
a = 2; b = 3; c = -14;
Δ = b2-4ac
Δ = 32-4·2·(-14)
Δ = 121
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{121}=11$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*2}=\frac{-14}{4} =-3+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*2}=\frac{8}{4} =2 $
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