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4n(n+8)=6
We move all terms to the left:
4n(n+8)-(6)=0
We multiply parentheses
4n^2+32n-6=0
a = 4; b = 32; c = -6;
Δ = b2-4ac
Δ = 322-4·4·(-6)
Δ = 1120
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{70}}{2*4}=\frac{-32-4\sqrt{70}}{8} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{70}}{2*4}=\frac{-32+4\sqrt{70}}{8} $
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