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2k^2+15k-6=0
a = 2; b = 15; c = -6;
Δ = b2-4ac
Δ = 152-4·2·(-6)
Δ = 273
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{-(15)-\sqrt{273}}{2*2}=\frac{-15-\sqrt{273}}{4} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{273}}{2*2}=\frac{-15+\sqrt{273}}{4} $
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