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4n^2+12n-280=0
a = 4; b = 12; c = -280;
Δ = b2-4ac
Δ = 122-4·4·(-280)
Δ = 4624
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{4624}=68$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-68}{2*4}=\frac{-80}{8} =-10 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+68}{2*4}=\frac{56}{8} =7 $
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