4=3.14r2

Solution for 4=3.14r2 equation:

4=3.14r^2

We move all terms to the left:

4-(3.14r^2)=0

We get rid of parentheses

-3.14r^2+4=0

a = -3.14; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-3.14)·4
Δ = 50.24
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{50.24}}{2*-3.14}=\frac{0-\sqrt{50.24}}{-6.28}
=-\frac{\sqrt{}}{-6.28}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{50.24}}{2*-3.14}=\frac{0+\sqrt{50.24}}{-6.28}
=\frac{\sqrt{}}{-6.28}
$