3y^2-3y-78=0

Solution for 3y^2-3y-78=0 equation:

3y^2-3y-78=0

a = 3; b = -3; c = -78;
Δ = b2-4ac
Δ = -32-4·3·(-78)
Δ = 945
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:

$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{105}}{2*3}=\frac{3-3\sqrt{105}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{105}}{2*3}=\frac{3+3\sqrt{105}}{6}
$