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n^2+n-600=0
a = 1; b = 1; c = -600;
Δ = b2-4ac
Δ = 12-4·1·(-600)
Δ = 2401
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{2401}=49$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-49}{2*1}=\frac{-50}{2} =-25 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+49}{2*1}=\frac{48}{2} =24 $
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