n^2-3n=48

Solution for n^2-3n=48 equation:

n^2-3n=48

We move all terms to the left:

n^2-3n-(48)=0

a = 1; b = -3; c = -48;
Δ = b2-4ac
Δ = -32-4·1·(-48)
Δ = 201
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{-(-3)-\sqrt{201}}{2*1}=\frac{3-\sqrt{201}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{201}}{2*1}=\frac{3+\sqrt{201}}{2}
$