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k^2+3k-248=0
a = 1; b = 3; c = -248;
Δ = b2-4ac
Δ = 32-4·1·(-248)
Δ = 1001
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{1001}}{2*1}=\frac{-3-\sqrt{1001}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1001}}{2*1}=\frac{-3+\sqrt{1001}}{2} $
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