-2(4y-3)-8y+6=4y(y-2)

Solution for -2(4y-3)-8y+6=4y(y-2) equation:

-2(4y-3)-8y+6=4y(y-2)

We move all terms to the left:

-2(4y-3)-8y+6-(4y(y-2))=0

We add all the numbers together, and all the variables

-8y-2(4y-3)-(4y(y-2))+6=0

We multiply parentheses

-8y-8y-(4y(y-2))+6+6=0


We calculate terms in parentheses: -(4y(y-2)), so:

4y(y-2)

We multiply parentheses

4y^2-8y

Back to the equation:

-(4y^2-8y)


We add all the numbers together, and all the variables

-16y-(4y^2-8y)+12=0

We get rid of parentheses

-4y^2-16y+8y+12=0

We add all the numbers together, and all the variables

-4y^2-8y+12=0

a = -4; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·(-4)·12
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*-4}=\frac{-8}{-8}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*-4}=\frac{24}{-8}
=-3
$