y-4y(y+2)=3y+4

Solution for y-4y(y+2)=3y+4 equation:

y-4y(y+2)=3y+4

We move all terms to the left:

y-4y(y+2)-(3y+4)=0

We multiply parentheses

-4y^2+y-8y-(3y+4)=0

We get rid of parentheses

-4y^2+y-8y-3y-4=0

We add all the numbers together, and all the variables

-4y^2-10y-4=0

a = -4; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·(-4)·(-4)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*-4}=\frac{4}{-8}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*-4}=\frac{16}{-8}
=-2
$