r2=0.16

Solution for r2=0.16 equation:

r2=0.16

We move all terms to the left:

r2-(0.16)=0

We add all the numbers together, and all the variables

r^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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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