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1950=5n^2-25n
We move all terms to the left:
1950-(5n^2-25n)=0
We get rid of parentheses
-5n^2+25n+1950=0
a = -5; b = 25; c = +1950;
Δ = b2-4ac
Δ = 252-4·(-5)·1950
Δ = 39625
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{39625}=\sqrt{25*1585}=\sqrt{25}*\sqrt{1585}=5\sqrt{1585}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{1585}}{2*-5}=\frac{-25-5\sqrt{1585}}{-10} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{1585}}{2*-5}=\frac{-25+5\sqrt{1585}}{-10} $
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