3X(n+1)=2X(n+5)

Solution for 3X(n+1)=2X(n+5) equation:

3X(X+1)=2X(X+5)

We move all terms to the left:

3X(X+1)-(2X(X+5))=0

We multiply parentheses

3X^2+3X-(2X(X+5))=0


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

2X(X+5)

We multiply parentheses

2X^2+10X

Back to the equation:

-(2X^2+10X)


We get rid of parentheses

3X^2-2X^2+3X-10X=0

We add all the numbers together, and all the variables

X^2-7X=0

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

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