(n-7)(2n+5)=0

Solution for (n-7)(2n+5)=0 equation:

(n-7)(2n+5)=0

We multiply parentheses ..

(+2n^2+5n-14n-35)=0

We get rid of parentheses

2n^2+5n-14n-35=0

We add all the numbers together, and all the variables

2n^2-9n-35=0

a = 2; b = -9; c = -35;
Δ = b2-4ac
Δ = -92-4·2·(-35)
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{-(-9)-19}{2*2}=\frac{-10}{4}
=-2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+19}{2*2}=\frac{28}{4}
=7
$