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n^2+9n-36=0
a = 1; b = 9; c = -36;
Δ = b2-4ac
Δ = 92-4·1·(-36)
Δ = 225
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{225}=15$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*1}=\frac{-24}{2} =-12 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*1}=\frac{6}{2} =3 $
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