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4n^2+28n+40=280^3
We move all terms to the left:
4n^2+28n+40-(280^3)=0
We add all the numbers together, and all the variables
4n^2+28n-21951960=0
a = 4; b = 28; c = -21951960;
Δ = b2-4ac
Δ = 282-4·4·(-21951960)
Δ = 351232144
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{351232144}=\sqrt{16*21952009}=\sqrt{16}*\sqrt{21952009}=4\sqrt{21952009}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{21952009}}{2*4}=\frac{-28-4\sqrt{21952009}}{8} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{21952009}}{2*4}=\frac{-28+4\sqrt{21952009}}{8} $
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