400p2=1

Solution for 400p2=1 equation:

400p^2=1

We move all terms to the left:

400p^2-(1)=0

a = 400; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·400·(-1)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*400}=\frac{-40}{800}
=-1/20
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*400}=\frac{40}{800}
=1/20
$