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n^2+19n-150=0
a = 1; b = 19; c = -150;
Δ = b2-4ac
Δ = 192-4·1·(-150)
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-31}{2*1}=\frac{-50}{2} =-25 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+31}{2*1}=\frac{12}{2} =6 $
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