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7k^2+51k-40=0
a = 7; b = 51; c = -40;
Δ = b2-4ac
Δ = 512-4·7·(-40)
Δ = 3721
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{3721}=61$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-61}{2*7}=\frac{-112}{14} =-8 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+61}{2*7}=\frac{10}{14} =5/7 $
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