3+r2=12

Solution for 3+r2=12 equation:

3+r2=12

We move all terms to the left:

3+r2-(12)=0

We add all the numbers together, and all the variables

r^2-9=0

a = 1; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·1·(-9)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*1}=\frac{-6}{2}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*1}=\frac{6}{2}
=3
$