3y^2-3y-100=0

Solution for 3y^2-3y-100=0 equation:

3y^2-3y-100=0

a = 3; b = -3; c = -100;
Δ = b2-4ac
Δ = -32-4·3·(-100)
Δ = 1209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{1209}}{2*3}=\frac{3-\sqrt{1209}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1209}}{2*3}=\frac{3+\sqrt{1209}}{6}
$