0=3y2+26y-35

Solution for 0=3y2+26y-35 equation:

0=3y^2+26y-35

We move all terms to the left:

0-(3y^2+26y-35)=0

We add all the numbers together, and all the variables

-(3y^2+26y-35)=0

We get rid of parentheses

-3y^2-26y+35=0

a = -3; b = -26; c = +35;
Δ = b2-4ac
Δ = -262-4·(-3)·35
Δ = 1096
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:

$\sqrt{\Delta}=\sqrt{1096}=\sqrt{4*274}=\sqrt{4}*\sqrt{274}=2\sqrt{274}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{274}}{2*-3}=\frac{26-2\sqrt{274}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{274}}{2*-3}=\frac{26+2\sqrt{274}}{-6}
$