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12n^2=4
We move all terms to the left:
12n^2-(4)=0
a = 12; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·12·(-4)
Δ = 192
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*12}=\frac{0-8\sqrt{3}}{24} =-\frac{8\sqrt{3}}{24} =-\frac{\sqrt{3}}{3} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*12}=\frac{0+8\sqrt{3}}{24} =\frac{8\sqrt{3}}{24} =\frac{\sqrt{3}}{3} $
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