3r^2-12+7=0

Solution for 3r^2-12+7=0 equation:

3r^2-12+7=0

We add all the numbers together, and all the variables

3r^2-5=0

a = 3; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·3·(-5)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*3}=\frac{0-2\sqrt{15}}{6}
=-\frac{2\sqrt{15}}{6}
=-\frac{\sqrt{15}}{3}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*3}=\frac{0+2\sqrt{15}}{6}
=\frac{2\sqrt{15}}{6}
=\frac{\sqrt{15}}{3}
$