A=3.14r^2+2r

Solution for A=3.14r^2+2r equation:

=3.14A^2+2A

We move all terms to the left:

-(3.14A^2+2A)=0

We get rid of parentheses

-3.14A^2-2A=0

a = -3.14; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-3.14)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-3.14}=\frac{0}{-6.28}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-3.14}=\frac{4}{-6.28}
=-2/3.14
$