3y-7y(y+5)=y-35

Solution for 3y-7y(y+5)=y-35 equation:

3y-7y(y+5)=y-35

We move all terms to the left:

3y-7y(y+5)-(y-35)=0

We multiply parentheses

-7y^2+3y-35y-(y-35)=0

We get rid of parentheses

-7y^2+3y-35y-y+35=0

We add all the numbers together, and all the variables

-7y^2-33y+35=0

a = -7; b = -33; c = +35;
Δ = b2-4ac
Δ = -332-4·(-7)·35
Δ = 2069
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{-(-33)-\sqrt{2069}}{2*-7}=\frac{33-\sqrt{2069}}{-14}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{2069}}{2*-7}=\frac{33+\sqrt{2069}}{-14}
$