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2n^2-9n-3=0
a = 2; b = -9; c = -3;
Δ = b2-4ac
Δ = -92-4·2·(-3)
Δ = 105
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{105}}{2*2}=\frac{9-\sqrt{105}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{105}}{2*2}=\frac{9+\sqrt{105}}{4} $
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