(4n-3)(n-1)=0

Solution for (4n-3)(n-1)=0 equation:

(4n-3)(n-1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4n^2-7n+3=0

a = 4; b = -7; c = +3;
Δ = b2-4ac
Δ = -72-4·4·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*4}=\frac{6}{8}
=3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*4}=\frac{8}{8}
=1
$