(18-r^2)+(25-r^2)=0

Solution for (18-r^2)+(25-r^2)=0 equation:

(18-r^2)+(25-r^2)=0

We get rid of parentheses

-r^2-r^2+18+25=0

We add all the numbers together, and all the variables

-2r^2+43=0

a = -2; b = 0; c = +43;
Δ = b2-4ac
Δ = 02-4·(-2)·43
Δ = 344
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:

$\sqrt{\Delta}=\sqrt{344}=\sqrt{4*86}=\sqrt{4}*\sqrt{86}=2\sqrt{86}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{86}}{2*-2}=\frac{0-2\sqrt{86}}{-4}
=-\frac{2\sqrt{86}}{-4}
=-\frac{\sqrt{86}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{86}}{2*-2}=\frac{0+2\sqrt{86}}{-4}
=\frac{2\sqrt{86}}{-4}
=\frac{\sqrt{86}}{-2}
$