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15k^2+7k-4=0
a = 15; b = 7; c = -4;
Δ = b2-4ac
Δ = 72-4·15·(-4)
Δ = 289
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{289}=17$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*15}=\frac{-24}{30} =-4/5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*15}=\frac{10}{30} =1/3 $
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