A(10)=3.14r2

Solution for A(10)=3.14r2 equation:

(10)=3.14A^2

We move all terms to the left:

(10)-(3.14A^2)=0

We get rid of parentheses

-3.14A^2+10=0

a = -3.14; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-3.14)·10
Δ = 125.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{125.6}}{2*-3.14}=\frac{0-\sqrt{125.6}}{-6.28}
=-\frac{\sqrt{}}{-6.28}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{125.6}}{2*-3.14}=\frac{0+\sqrt{125.6}}{-6.28}
=\frac{\sqrt{}}{-6.28}
$