(3n+2)(n=3)

Solution for (3n+2)(n=3) equation:

(3n+2)(n=3)

We move all terms to the left:

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

We multiply parentheses ..

(+3n^2-9n+2n-6)=0

We get rid of parentheses

3n^2-9n+2n-6=0

We add all the numbers together, and all the variables

3n^2-7n-6=0

a = 3; b = -7; c = -6;
Δ = b2-4ac
Δ = -72-4·3·(-6)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-11}{2*3}=\frac{-4}{6}
=-2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+11}{2*3}=\frac{18}{6}
=3
$