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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

3n+4n-((+5n)n-2)-2=0

We add all the numbers together, and all the variables

7n-((+5n)n-2)-2=0


We calculate terms in parentheses: -((+5n)n-2), so:

(+5n)n-2

We multiply parentheses

5n^2-2

Back to the equation:

-(5n^2-2)


We get rid of parentheses

-5n^2+7n+2-2=0

We add all the numbers together, and all the variables

-5n^2+7n=0

a = -5; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-5)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-5}=\frac{-14}{-10}
=1+2/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-5}=\frac{0}{-10}
=0
$