3y^2-24y-41=0

Solution for 3y^2-24y-41=0 equation:

3y^2-24y-41=0

a = 3; b = -24; c = -41;
Δ = b2-4ac
Δ = -242-4·3·(-41)
Δ = 1068
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:

$\sqrt{\Delta}=\sqrt{1068}=\sqrt{4*267}=\sqrt{4}*\sqrt{267}=2\sqrt{267}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-2\sqrt{267}}{2*3}=\frac{24-2\sqrt{267}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+2\sqrt{267}}{2*3}=\frac{24+2\sqrt{267}}{6}
$