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n(n+1)=240
We move all terms to the left:
n(n+1)-(240)=0
We multiply parentheses
n^2+n-240=0
a = 1; b = 1; c = -240;
Δ = b2-4ac
Δ = 12-4·1·(-240)
Δ = 961
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{961}=31$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*1}=\frac{-32}{2} =-16 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*1}=\frac{30}{2} =15 $
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