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4k^2+k-138=0
a = 4; b = 1; c = -138;
Δ = b2-4ac
Δ = 12-4·4·(-138)
Δ = 2209
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{2209}=47$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-47}{2*4}=\frac{-48}{8} =-6 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+47}{2*4}=\frac{46}{8} =5+3/4 $
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