1p2+0.04p-0.16=0

Solution for 1p2+0.04p-0.16=0 equation:

1p^2+0.04p-0.16=0

We add all the numbers together, and all the variables

p^2+0.04p-0.16=0

a = 1; b = 0.04; c = -0.16;
Δ = b2-4ac
Δ = 0.042-4·1·(-0.16)
Δ = 0.6416
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{-(0.04)-\sqrt{0.6416}}{2*1}=\frac{-0.04-\sqrt{0.6416}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.04)+\sqrt{0.6416}}{2*1}=\frac{-0.04+\sqrt{0.6416}}{2}
$