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5n^2-45n+100=0
a = 5; b = -45; c = +100;
Δ = b2-4ac
Δ = -452-4·5·100
Δ = 25
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{25}=5$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-5}{2*5}=\frac{40}{10} =4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+5}{2*5}=\frac{50}{10} =5 $
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