140=n^2-4n

Solution for 140=n^2-4n equation:

140=n^2-4n

We move all terms to the left:

140-(n^2-4n)=0

We get rid of parentheses

-n^2+4n+140=0

We add all the numbers together, and all the variables

-1n^2+4n+140=0

a = -1; b = 4; c = +140;
Δ = b2-4ac
Δ = 42-4·(-1)·140
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-24}{2*-1}=\frac{-28}{-2}
=+14
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+24}{2*-1}=\frac{20}{-2}
=-10
$