(5)r=3.14r^2

Solution for (5)r=3.14r^2 equation:

(5)r=3.14r^2

We move all terms to the left:

(5)r-(3.14r^2)=0

We get rid of parentheses

-3.14r^2+5r=0

a = -3.14; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-3.14)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-3.14}=\frac{-10}{-6.28}
=1+1/1.6881720430108
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-3.14}=\frac{0}{-6.28}
=0
$