n2+42n=0

Solution for n2+42n=0 equation:

n2+42n=0

We add all the numbers together, and all the variables

n^2+42n=0

a = 1; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·1·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*1}=\frac{-84}{2}
=-42
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*1}=\frac{0}{2}
=0
$