(240.5)2+(r-46)2=r2

Solution for (240.5)2+(r-46)2=r2 equation:

(240.5)2+(r-46)2=r2

We move all terms to the left:

(240.5)2+(r-46)2-(r2)=0

We add all the numbers together, and all the variables

-1r^2+(r-46)2+481=0

We multiply parentheses

-1r^2+2r-92+481=0

We add all the numbers together, and all the variables

-1r^2+2r+389=0

a = -1; b = 2; c = +389;
Δ = b2-4ac
Δ = 22-4·(-1)·389
Δ = 1560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1560}=\sqrt{4*390}=\sqrt{4}*\sqrt{390}=2\sqrt{390}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{390}}{2*-1}=\frac{-2-2\sqrt{390}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{390}}{2*-1}=\frac{-2+2\sqrt{390}}{-2}
$