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12n^2-2n-5=0
a = 12; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·12·(-5)
Δ = 244
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{61}}{2*12}=\frac{2-2\sqrt{61}}{24} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{61}}{2*12}=\frac{2+2\sqrt{61}}{24} $
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