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n2+6n+5=8
We move all terms to the left:
n2+6n+5-(8)=0
We add all the numbers together, and all the variables
n^2+6n-3=0
a = 1; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·1·(-3)
Δ = 48
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*1}=\frac{-6-4\sqrt{3}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*1}=\frac{-6+4\sqrt{3}}{2} $
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