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

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

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

We move all terms to the left:

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

We multiply parentheses

8y-(3(y+8)y)-16=0


We calculate terms in parentheses: -(3(y+8)y), so:

3(y+8)y

We multiply parentheses

3y^2+24y

Back to the equation:

-(3y^2+24y)


We get rid of parentheses

-3y^2+8y-24y-16=0

We add all the numbers together, and all the variables

-3y^2-16y-16=0

a = -3; b = -16; c = -16;
Δ = b2-4ac
Δ = -162-4·(-3)·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*-3}=\frac{8}{-6}
=-1+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*-3}=\frac{24}{-6}
=-4
$