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2n^2-3n-119=0
a = 2; b = -3; c = -119;
Δ = b2-4ac
Δ = -32-4·2·(-119)
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-31}{2*2}=\frac{-28}{4} =-7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+31}{2*2}=\frac{34}{4} =8+1/2 $
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