A(2)A(r)=3.14r^2

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

(2)(A)=3.14A^2

We move all terms to the left:

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