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12n^2+13n+3=0
a = 12; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·12·3
Δ = 25
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*12}=\frac{-18}{24} =-3/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*12}=\frac{-8}{24} =-1/3 $
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