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2k^2-21k-36=0
a = 2; b = -21; c = -36;
Δ = b2-4ac
Δ = -212-4·2·(-36)
Δ = 729
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{729}=27$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-27}{2*2}=\frac{-6}{4} =-1+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+27}{2*2}=\frac{48}{4} =12 $
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