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

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

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

We multiply parentheses ..

(+2n^2-10n+5n-25)=0

We get rid of parentheses

2n^2-10n+5n-25=0

We add all the numbers together, and all the variables

2n^2-5n-25=0

a = 2; b = -5; c = -25;
Δ = b2-4ac
Δ = -52-4·2·(-25)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-15}{2*2}=\frac{-10}{4}
=-2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+15}{2*2}=\frac{20}{4}
=5
$