25n2+4=20

Solution for 25n2+4=20 equation:

25n^2+4=20

We move all terms to the left:

25n^2+4-(20)=0

We add all the numbers together, and all the variables

25n^2-16=0

a = 25; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·25·(-16)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*25}=\frac{-40}{50}
=-4/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*25}=\frac{40}{50}
=4/5
$