(2n+5)(2n+9)=0

Solution for (2n+5)(2n+9)=0 equation:

(2n+5)(2n+9)=0

We multiply parentheses ..

(+4n^2+18n+10n+45)=0

We get rid of parentheses

4n^2+18n+10n+45=0

We add all the numbers together, and all the variables

4n^2+28n+45=0

a = 4; b = 28; c = +45;
Δ = b2-4ac
Δ = 282-4·4·45
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8}{2*4}=\frac{-36}{8}
=-4+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8}{2*4}=\frac{-20}{8}
=-2+1/2
$