16r+r^2*3.14=32

Solution for 16r+r^2*3.14=32 equation:

16r+r^2*3.14=32

We move all terms to the left:

16r+r^2*3.14-(32)=0

Wy multiply elements

3.14r^2+16r-32=0

a = 3.14; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·3.14·(-32)
Δ = 657.92
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{-(16)-\sqrt{657.92}}{2*3.14}=\frac{-16-\sqrt{657.92}}{6.28}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+\sqrt{657.92}}{2*3.14}=\frac{-16+\sqrt{657.92}}{6.28}
$