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2n^2+15n+15=0
a = 2; b = 15; c = +15;
Δ = b2-4ac
Δ = 152-4·2·15
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{-(15)-\sqrt{105}}{2*2}=\frac{-15-\sqrt{105}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{105}}{2*2}=\frac{-15+\sqrt{105}}{4} $
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