4r^2=12+8r

Solution for 4r^2=12+8r equation:

4r^2=12+8r

We move all terms to the left:

4r^2-(12+8r)=0

We add all the numbers together, and all the variables

4r^2-(8r+12)=0

We get rid of parentheses

4r^2-8r-12=0

a = 4; b = -8; c = -12;
Δ = b2-4ac
Δ = -82-4·4·(-12)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*4}=\frac{-8}{8}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*4}=\frac{24}{8}
=3
$