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8n(4n+5)=7
We move all terms to the left:
8n(4n+5)-(7)=0
We multiply parentheses
32n^2+40n-7=0
a = 32; b = 40; c = -7;
Δ = b2-4ac
Δ = 402-4·32·(-7)
Δ = 2496
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{39}}{2*32}=\frac{-40-8\sqrt{39}}{64} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{39}}{2*32}=\frac{-40+8\sqrt{39}}{64} $
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