1n^2+4n=140

Solution for 1n^2+4n=140 equation:

1n^2+4n=140

We move all terms to the left:

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

We add all the numbers together, and all the variables

n^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
$