3r^2=-48-32r

Solution for 3r^2=-48-32r equation:

3r^2=-48-32r

We move all terms to the left:

3r^2-(-48-32r)=0

We add all the numbers together, and all the variables

3r^2-(-32r-48)=0

We get rid of parentheses

3r^2+32r+48=0

a = 3; b = 32; c = +48;
Δ = b2-4ac
Δ = 322-4·3·48
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{7}}{2*3}=\frac{-32-8\sqrt{7}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{7}}{2*3}=\frac{-32+8\sqrt{7}}{6}
$