(3y^2)-14y+7=0

Solution for (3y^2)-14y+7=0 equation:

(3y^2)-14y+7=0

a = 3; b = -14; c = +7;
Δ = b2-4ac
Δ = -142-4·3·7
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{7}}{2*3}=\frac{14-4\sqrt{7}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{7}}{2*3}=\frac{14+4\sqrt{7}}{6}
$