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23n^2+45n-2=0
a = 23; b = 45; c = -2;
Δ = b2-4ac
Δ = 452-4·23·(-2)
Δ = 2209
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{2209}=47$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-47}{2*23}=\frac{-92}{46} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+47}{2*23}=\frac{2}{46} =1/23 $
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