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7k^2+29k-30=0
a = 7; b = 29; c = -30;
Δ = b2-4ac
Δ = 292-4·7·(-30)
Δ = 1681
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{1681}=41$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*7}=\frac{-70}{14} =-5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*7}=\frac{12}{14} =6/7 $
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