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n^2+23n-1200=0
a = 1; b = 23; c = -1200;
Δ = b2-4ac
Δ = 232-4·1·(-1200)
Δ = 5329
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{5329}=73$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-73}{2*1}=\frac{-96}{2} =-48 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+73}{2*1}=\frac{50}{2} =25 $
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