(4)7=(n+2)n

Solution for (4)7=(n+2)n equation:

(4)7=(n+2)n

We move all terms to the left:

(4)7-((n+2)n)=0


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

(n+2)n

We multiply parentheses

n^2+2n

Back to the equation:

-(n^2+2n)


We get rid of parentheses

-n^2-2n+47=0

We add all the numbers together, and all the variables

-1n^2-2n+47=0

a = -1; b = -2; c = +47;
Δ = b2-4ac
Δ = -22-4·(-1)·47
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8\sqrt{3}}{2*-1}=\frac{2-8\sqrt{3}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8\sqrt{3}}{2*-1}=\frac{2+8\sqrt{3}}{-2}
$