34+8=n(n+4)

Solution for 34+8=n(n+4) equation:

34+8=n(n+4)

We move all terms to the left:

34+8-(n(n+4))=0

We add all the numbers together, and all the variables

-(n(n+4))+42=0


We calculate terms in parentheses: -(n(n+4)), so:

n(n+4)

We multiply parentheses

n^2+4n

Back to the equation:

-(n^2+4n)


We get rid of parentheses

-n^2-4n+42=0

We add all the numbers together, and all the variables

-1n^2-4n+42=0

a = -1; b = -4; c = +42;
Δ = b2-4ac
Δ = -42-4·(-1)·42
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{46}}{2*-1}=\frac{4-2\sqrt{46}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{46}}{2*-1}=\frac{4+2\sqrt{46}}{-2}
$