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42k^2-19k=0
a = 42; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·42·0
Δ = 361
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{361}=19$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*42}=\frac{0}{84} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*42}=\frac{38}{84} =19/42 $
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