12y^2+y=35=0

Solution for 12y^2+y=35=0 equation:

12y^2+y=35=0

We move all terms to the left:

12y^2+y-(35)=0

a = 12; b = 1; c = -35;
Δ = b2-4ac
Δ = 12-4·12·(-35)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*12}=\frac{-42}{24}
=-1+3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*12}=\frac{40}{24}
=1+2/3
$