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2k^2-21k+20=0
a = 2; b = -21; c = +20;
Δ = b2-4ac
Δ = -212-4·2·20
Δ = 281
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{-(-21)-\sqrt{281}}{2*2}=\frac{21-\sqrt{281}}{4} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{281}}{2*2}=\frac{21+\sqrt{281}}{4} $
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