9=r2+3

Solution for 9=r2+3 equation:

9=r2+3

We move all terms to the left:

9-(r2+3)=0

We add all the numbers together, and all the variables

-(+r^2+3)+9=0

We get rid of parentheses

-r^2-3+9=0

We add all the numbers together, and all the variables

-1r^2+6=0

a = -1; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-1)·6
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-1}=\frac{0-2\sqrt{6}}{-2}
=-\frac{2\sqrt{6}}{-2}
=-\frac{\sqrt{6}}{-1}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-1}=\frac{0+2\sqrt{6}}{-2}
=\frac{2\sqrt{6}}{-2}
=\frac{\sqrt{6}}{-1}
$