p^2=(0.4)^2

Solution for p^2=(0.4)^2 equation:

p^2=(0.4)^2

We move all terms to the left:

p^2-((0.4)^2)=0

We add all the numbers together, and all the variables

p^2-0.16=0

a = 1; b = 0; c = -0.16;
Δ = b2-4ac
Δ = 02-4·1·(-0.16)
Δ = 0.64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.64}}{2*1}=\frac{0-\sqrt{0.64}}{2}
=-\frac{\sqrt{}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.64}}{2*1}=\frac{0+\sqrt{0.64}}{2}
=\frac{\sqrt{}}{2}
$