(4n-3)(n=2)

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

(4n-3)(n=2)

We move all terms to the left:

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

We multiply parentheses ..

(+4n^2-8n-3n+6)=0

We get rid of parentheses

4n^2-8n-3n+6=0

We add all the numbers together, and all the variables

4n^2-11n+6=0

a = 4; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·4·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*4}=\frac{6}{8}
=3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*4}=\frac{16}{8}
=2
$