16=3.14(r)^2

Solution for 16=3.14(r)^2 equation:

16=3.14(r)^2

We move all terms to the left:

16-(3.14(r)^2)=0

We get rid of parentheses

-3.14r^2+16=0

a = -3.14; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-3.14)·16
Δ = 200.96
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{200.96}}{2*-3.14}=\frac{0-\sqrt{200.96}}{-6.28}
=-\frac{\sqrt{}}{-6.28}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{200.96}}{2*-3.14}=\frac{0+\sqrt{200.96}}{-6.28}
=\frac{\sqrt{}}{-6.28}
$