n2+4=29n=

Solution for n2+4=29n= equation:

n2+4=29n=

We move all terms to the left:

n2+4-(29n)=0

We add all the numbers together, and all the variables

n^2-29n+4=0

a = 1; b = -29; c = +4;
Δ = b2-4ac
Δ = -292-4·1·4
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-5\sqrt{33}}{2*1}=\frac{29-5\sqrt{33}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+5\sqrt{33}}{2*1}=\frac{29+5\sqrt{33}}{2}
$