n=43+(n-47/4)

Solution for n=43+(n-47/4) equation:

n=43+(n-47/4)

We move all terms to the left:

n-(43+(n-47/4))=0

We add all the numbers together, and all the variables

n-(43+(+n-47/4))=0

We multiply all the terms by the denominator


n*4))-(43+(+n-47=0

We add all the numbers together, and all the variables

n+n*4))-(43+(-47=0

Wy multiply elements

4n^2+n-47=0

a = 4; b = 1; c = -47;
Δ = b2-4ac
Δ = 12-4·4·(-47)
Δ = 753
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{-(1)-\sqrt{753}}{2*4}=\frac{-1-\sqrt{753}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{753}}{2*4}=\frac{-1+\sqrt{753}}{8}
$