(4n+3)(2n-3)=0

Solution for (4n+3)(2n-3)=0 equation:

(4n+3)(2n-3)=0

We multiply parentheses ..

(+8n^2-12n+6n-9)=0

We get rid of parentheses

8n^2-12n+6n-9=0

We add all the numbers together, and all the variables

8n^2-6n-9=0

a = 8; b = -6; c = -9;
Δ = b2-4ac
Δ = -62-4·8·(-9)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*8}=\frac{-12}{16}
=-3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*8}=\frac{24}{16}
=1+1/2
$