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

Solution for 3n+4=5(n+2)2n equation:

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

We move all terms to the left:

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


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

5(n+2)2n

We multiply parentheses

10n^2+20n

Back to the equation:

-(10n^2+20n)


We get rid of parentheses

-10n^2+3n-20n+4=0

We add all the numbers together, and all the variables

-10n^2-17n+4=0

a = -10; b = -17; c = +4;
Δ = b2-4ac
Δ = -172-4·(-10)·4
Δ = 449
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{449}}{2*-10}=\frac{17-\sqrt{449}}{-20}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{449}}{2*-10}=\frac{17+\sqrt{449}}{-20}
$