3y^2-48y+1=0

Solution for 3y^2-48y+1=0 equation:

3y^2-48y+1=0

a = 3; b = -48; c = +1;
Δ = b2-4ac
Δ = -482-4·3·1
Δ = 2292
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:

$\sqrt{\Delta}=\sqrt{2292}=\sqrt{4*573}=\sqrt{4}*\sqrt{573}=2\sqrt{573}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-2\sqrt{573}}{2*3}=\frac{48-2\sqrt{573}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+2\sqrt{573}}{2*3}=\frac{48+2\sqrt{573}}{6}
$