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23n^2+9n=0
a = 23; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·23·0
Δ = 81
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{81}=9$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*23}=\frac{-18}{46} =-9/23 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*23}=\frac{0}{46} =0 $
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