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(4n+1)(2n-5)=0
We multiply parentheses ..
(+8n^2-20n+2n-5)=0
We get rid of parentheses
8n^2-20n+2n-5=0
We add all the numbers together, and all the variables
8n^2-18n-5=0
a = 8; b = -18; c = -5;
Δ = b2-4ac
Δ = -182-4·8·(-5)
Δ = 484
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{484}=22$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*8}=\frac{-4}{16} =-1/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*8}=\frac{40}{16} =2+1/2 $
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