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5n^2-19n-30=0
a = 5; b = -19; c = -30;
Δ = b2-4ac
Δ = -192-4·5·(-30)
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-31}{2*5}=\frac{-12}{10} =-1+1/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+31}{2*5}=\frac{50}{10} =5 $
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