3r^2-12r+7=0

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

3r^2-12r+7=0

a = 3; b = -12; c = +7;
Δ = b2-4ac
Δ = -122-4·3·7
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{-(-12)-2\sqrt{15}}{2*3}=\frac{12-2\sqrt{15}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{15}}{2*3}=\frac{12+2\sqrt{15}}{6}
$