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2n^2+2n-1200=0
a = 2; b = 2; c = -1200;
Δ = b2-4ac
Δ = 22-4·2·(-1200)
Δ = 9604
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{9604}=98$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-98}{2*2}=\frac{-100}{4} =-25 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+98}{2*2}=\frac{96}{4} =24 $
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