n^2-38=n+4

Solution for n^2-38=n+4 equation:

n^2-38=n+4

We move all terms to the left:

n^2-38-(n+4)=0

We get rid of parentheses

n^2-n-4-38=0

We add all the numbers together, and all the variables

n^2-1n-42=0

a = 1; b = -1; c = -42;
Δ = b2-4ac
Δ = -12-4·1·(-42)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-13}{2*1}=\frac{-12}{2}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+13}{2*1}=\frac{14}{2}
=7
$