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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(y(y+1)-4), so:

y(y+1)-4

We multiply parentheses

y^2+y-4

Back to the equation:

-(y^2+y-4)


We add all the numbers together, and all the variables

-1y-(y^2+y-4)-4=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1y^2-2y=0

a = -1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-1}=\frac{0}{-2}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-1}=\frac{4}{-2}
=-2
$