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k^2+15k-56=0
a = 1; b = 15; c = -56;
Δ = b2-4ac
Δ = 152-4·1·(-56)
Δ = 449
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{-(15)-\sqrt{449}}{2*1}=\frac{-15-\sqrt{449}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{449}}{2*1}=\frac{-15+\sqrt{449}}{2} $
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