2-4(y-1)=5y+(y-3);y=3

Solution for 2-4(y-1)=5y+(y-3);y=3 equation:

2-4(y-1)=5y+(y-3)y=3

We move all terms to the left:

2-4(y-1)-(5y+(y-3)y)=0

We multiply parentheses

-4y-(5y+(y-3)y)+4+2=0


We calculate terms in parentheses: -(5y+(y-3)y), so:

5y+(y-3)y

We multiply parentheses

y^2+5y-3y

We add all the numbers together, and all the variables

y^2+2y

Back to the equation:

-(y^2+2y)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-y^2-4y-2y+6=0

We add all the numbers together, and all the variables

-1y^2-6y+6=0

a = -1; b = -6; c = +6;
Δ = b2-4ac
Δ = -62-4·(-1)·6
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*-1}=\frac{6-2\sqrt{15}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*-1}=\frac{6+2\sqrt{15}}{-2}
$