3r^2-16-12=0

Solution for 3r^2-16-12=0 equation:

3r^2-16-12=0

We add all the numbers together, and all the variables

3r^2-28=0

a = 3; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·3·(-28)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*3}=\frac{0-4\sqrt{21}}{6}
=-\frac{4\sqrt{21}}{6}
=-\frac{2\sqrt{21}}{3}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*3}=\frac{0+4\sqrt{21}}{6}
=\frac{4\sqrt{21}}{6}
=\frac{2\sqrt{21}}{3}
$