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2n^2+5n-25=0
a = 2; b = 5; c = -25;
Δ = b2-4ac
Δ = 52-4·2·(-25)
Δ = 225
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{225}=15$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-15}{2*2}=\frac{-20}{4} =-5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+15}{2*2}=\frac{10}{4} =2+1/2 $
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