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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(y-4)(y-2)+(y-4)(y+2)

We multiply parentheses ..

(+y^2-2y-4y+8)+(y-4)(y+2)

We get rid of parentheses

y^2-2y-4y+(y-4)(y+2)+8

We multiply parentheses ..

y^2+(+y^2+2y-4y-8)-2y-4y+8

We add all the numbers together, and all the variables

y^2+(+y^2+2y-4y-8)-6y+8

We get rid of parentheses

y^2+y^2+2y-4y-6y-8+8

We add all the numbers together, and all the variables

2y^2-8y

Back to the equation:

-(2y^2-8y)


We get rid of parentheses

y^2-2y^2-3y+2y+8y-6=0

We add all the numbers together, and all the variables

-1y^2+7y-6=0

a = -1; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·(-1)·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*-1}=\frac{-12}{-2}
=+6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*-1}=\frac{-2}{-2}
=1
$