4n2=125-n2

Solution for 4n2=125-n2 equation:

4n^2=125-n2

We move all terms to the left:

4n^2-(125-n2)=0

We add all the numbers together, and all the variables

4n^2-(-1n^2+125)=0

We get rid of parentheses

4n^2+1n^2-125=0

We add all the numbers together, and all the variables

5n^2-125=0

a = 5; b = 0; c = -125;
Δ = b2-4ac
Δ = 02-4·5·(-125)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-50}{2*5}=\frac{-50}{10}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+50}{2*5}=\frac{50}{10}
=5
$