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7n^2+2n=8
We move all terms to the left:
7n^2+2n-(8)=0
a = 7; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·7·(-8)
Δ = 228
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{57}}{2*7}=\frac{-2-2\sqrt{57}}{14} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{57}}{2*7}=\frac{-2+2\sqrt{57}}{14} $
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