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45n^2+10n=0
a = 45; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·45·0
Δ = 100
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{100}=10$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*45}=\frac{-20}{90} =-2/9 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*45}=\frac{0}{90} =0 $
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