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2n^2+12n+16n=0
We add all the numbers together, and all the variables
2n^2+28n=0
a = 2; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·2·0
Δ = 784
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{784}=28$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*2}=\frac{-56}{4} =-14 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*2}=\frac{0}{4} =0 $
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