57840.25+(r-46)2=r2

Solution for 57840.25+(r-46)2=r2 equation:

57840.25+(r-46)2=r2

We move all terms to the left:

57840.25+(r-46)2-(r2)=0

We add all the numbers together, and all the variables

-1r^2+(r-46)2+57840.25=0

We multiply parentheses

-1r^2+2r-92+57840.25=0

We add all the numbers together, and all the variables

-1r^2+2r+57748.25=0

a = -1; b = 2; c = +57748.25;
Δ = b2-4ac
Δ = 22-4·(-1)·57748.25
Δ = 230997
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{-(2)-\sqrt{230997}}{2*-1}=\frac{-2-\sqrt{230997}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{230997}}{2*-1}=\frac{-2+\sqrt{230997}}{-2}
$