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5n^2+n-6=0
a = 5; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·5·(-6)
Δ = 121
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{121}=11$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*5}=\frac{-12}{10} =-1+1/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*5}=\frac{10}{10} =1 $
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