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

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

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

We multiply parentheses ..

(+2n^2+10n-9n-45)=0

We get rid of parentheses

2n^2+10n-9n-45=0

We add all the numbers together, and all the variables

2n^2+n-45=0

a = 2; b = 1; c = -45;
Δ = b2-4ac
Δ = 12-4·2·(-45)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*2}=\frac{-20}{4}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*2}=\frac{18}{4}
=4+1/2
$