(n-7)(3n+1)=0

Solution for (n-7)(3n+1)=0 equation:

(n-7)(3n+1)=0

We multiply parentheses ..

(+3n^2+n-21n-7)=0

We get rid of parentheses

3n^2+n-21n-7=0

We add all the numbers together, and all the variables

3n^2-20n-7=0

a = 3; b = -20; c = -7;
Δ = b2-4ac
Δ = -202-4·3·(-7)
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{-(-20)-22}{2*3}=\frac{-2}{6}
=-1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+22}{2*3}=\frac{42}{6}
=7
$