n+n+1+n2=48

Solution for n+n+1+n2=48 equation:

n+n+1+n2=48

We move all terms to the left:

n+n+1+n2-(48)=0

We add all the numbers together, and all the variables

n^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}
$