(n-3)(4n+3)=0

Solution for (n-3)(4n+3)=0 equation:

(n-3)(4n+3)=0

We multiply parentheses ..

(+4n^2+3n-12n-9)=0

We get rid of parentheses

4n^2+3n-12n-9=0

We add all the numbers together, and all the variables

4n^2-9n-9=0

a = 4; b = -9; c = -9;
Δ = b2-4ac
Δ = -92-4·4·(-9)
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{-(-9)-15}{2*4}=\frac{-6}{8}
=-3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*4}=\frac{24}{8}
=3
$