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45n^2-5=0
a = 45; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·45·(-5)
Δ = 900
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{900}=30$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*45}=\frac{-30}{90} =-1/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*45}=\frac{30}{90} =1/3 $
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