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4n^2+8n-2=0
a = 4; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·4·(-2)
Δ = 96
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{6}}{2*4}=\frac{-8-4\sqrt{6}}{8} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{6}}{2*4}=\frac{-8+4\sqrt{6}}{8} $
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