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(4n+5)(5n+1)=0
We multiply parentheses ..
(+20n^2+4n+25n+5)=0
We get rid of parentheses
20n^2+4n+25n+5=0
We add all the numbers together, and all the variables
20n^2+29n+5=0
a = 20; b = 29; c = +5;
Δ = b2-4ac
Δ = 292-4·20·5
Δ = 441
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{441}=21$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*20}=\frac{-50}{40} =-1+1/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*20}=\frac{-8}{40} =-1/5 $
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