3y^2-96+y=90

Solution for 3y^2-96+y=90 equation:

3y^2-96+y=90

We move all terms to the left:

3y^2-96+y-(90)=0

We add all the numbers together, and all the variables

3y^2+y-186=0

a = 3; b = 1; c = -186;
Δ = b2-4ac
Δ = 12-4·3·(-186)
Δ = 2233
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{-(1)-\sqrt{2233}}{2*3}=\frac{-1-\sqrt{2233}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2233}}{2*3}=\frac{-1+\sqrt{2233}}{6}
$