(Y-4)(y+3)(2)+(y+5)(y+3)(4)=(y-4)(y+5)(6)

Solution for (Y-4)(y+3)(2)+(y+5)(y+3)(4)=(y-4)(y+5)(6) equation:

(-4)(Y+3)(2)+(Y+5)(Y+3)(4)=(Y-4)(Y+5)(6)

We move all terms to the left:

(-4)(Y+3)(2)+(Y+5)(Y+3)(4)-((Y-4)(Y+5)(6))=0

We multiply parentheses ..

(-4Y-12)2+(Y+5)(Y+3)4-((Y-4)(Y+5)6)=0


We calculate terms in parentheses: -((Y-4)(Y+5)6), so:

(Y-4)(Y+5)6

We multiply parentheses ..

(+Y^2+5Y-4Y-20)6

We multiply parentheses

6Y^2+30Y-24Y-120

We add all the numbers together, and all the variables

6Y^2+6Y-120

Back to the equation:

-(6Y^2+6Y-120)


We multiply parentheses

-8Y+(Y+5)(Y+3)4-(6Y^2+6Y-120)-24=0

We get rid of parentheses

-6Y^2-8Y+(Y+5)(Y+3)4-6Y+120-24=0

We multiply parentheses ..

-6Y^2+(+Y^2+3Y+5Y+15)4-8Y-6Y+120-24=0

We add all the numbers together, and all the variables

-6Y^2+(+Y^2+3Y+5Y+15)4-14Y+96=0

We multiply parentheses

-6Y^2+4Y^2+12Y+20Y-14Y+60+96=0

We add all the numbers together, and all the variables

-2Y^2+18Y+156=0

a = -2; b = 18; c = +156;
Δ = b2-4ac
Δ = 182-4·(-2)·156
Δ = 1572
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:

$\sqrt{\Delta}=\sqrt{1572}=\sqrt{4*393}=\sqrt{4}*\sqrt{393}=2\sqrt{393}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{393}}{2*-2}=\frac{-18-2\sqrt{393}}{-4}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{393}}{2*-2}=\frac{-18+2\sqrt{393}}{-4}
$