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2n(2n+2)=168
We move all terms to the left:
2n(2n+2)-(168)=0
We multiply parentheses
4n^2+4n-168=0
a = 4; b = 4; c = -168;
Δ = b2-4ac
Δ = 42-4·4·(-168)
Δ = 2704
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{2704}=52$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-52}{2*4}=\frac{-56}{8} =-7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+52}{2*4}=\frac{48}{8} =6 $
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