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-2k^2+6k-4=0
a = -2; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·(-2)·(-4)
Δ = 4
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{4}=2$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-2}=\frac{-8}{-4} =+2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-2}=\frac{-4}{-4} =1 $
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