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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(2y+(y-3)y), so:

2y+(y-3)y

We multiply parentheses

y^2+2y-3y

We add all the numbers together, and all the variables

y^2-1y

Back to the equation:

-(y^2-1y)


We add all the numbers together, and all the variables

-5y-(y^2-1y)+9=0

We get rid of parentheses

-y^2-5y+1y+9=0

We add all the numbers together, and all the variables

-1y^2-4y+9=0

a = -1; b = -4; c = +9;
Δ = b2-4ac
Δ = -42-4·(-1)·9
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{13}}{2*-1}=\frac{4-2\sqrt{13}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{13}}{2*-1}=\frac{4+2\sqrt{13}}{-2}
$