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k^2+5k-300=0
a = 1; b = 5; c = -300;
Δ = b2-4ac
Δ = 52-4·1·(-300)
Δ = 1225
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{1225}=35$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-35}{2*1}=\frac{-40}{2} =-20 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+35}{2*1}=\frac{30}{2} =15 $
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