8y+4(2+y)=3(y-4)10y

Solution for 8y+4(2+y)=3(y-4)10y equation:

8y+4(2+y)=3(y-4)10y

We move all terms to the left:

8y+4(2+y)-(3(y-4)10y)=0

We add all the numbers together, and all the variables

8y+4(y+2)-(3(y-4)10y)=0

We multiply parentheses

8y+4y-(3(y-4)10y)+8=0


We calculate terms in parentheses: -(3(y-4)10y), so:

3(y-4)10y

We multiply parentheses

30y^2-120y

Back to the equation:

-(30y^2-120y)


We add all the numbers together, and all the variables

12y-(30y^2-120y)+8=0

We get rid of parentheses

-30y^2+12y+120y+8=0

We add all the numbers together, and all the variables

-30y^2+132y+8=0

a = -30; b = 132; c = +8;
Δ = b2-4ac
Δ = 1322-4·(-30)·8
Δ = 18384
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:

$\sqrt{\Delta}=\sqrt{18384}=\sqrt{16*1149}=\sqrt{16}*\sqrt{1149}=4\sqrt{1149}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(132)-4\sqrt{1149}}{2*-30}=\frac{-132-4\sqrt{1149}}{-60}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(132)+4\sqrt{1149}}{2*-30}=\frac{-132+4\sqrt{1149}}{-60}
$