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2n^2+4n-6=0
a = 2; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·2·(-6)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*2}=\frac{-12}{4} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*2}=\frac{4}{4} =1 $
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