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40=n(2n+3)
We move all terms to the left:
40-(n(2n+3))=0
We calculate terms in parentheses: -(n(2n+3)), so:We get rid of parentheses
n(2n+3)
We multiply parentheses
2n^2+3n
Back to the equation:
-(2n^2+3n)
-2n^2-3n+40=0
a = -2; b = -3; c = +40;
Δ = b2-4ac
Δ = -32-4·(-2)·40
Δ = 329
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{-(-3)-\sqrt{329}}{2*-2}=\frac{3-\sqrt{329}}{-4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{329}}{2*-2}=\frac{3+\sqrt{329}}{-4} $
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