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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(Y(Y-2)+6), so:

Y(Y-2)+6

We multiply parentheses

Y^2-2Y+6

Back to the equation:

-(Y^2-2Y+6)


We get rid of parentheses

-Y^2+3Y+2Y-6+12=0

We add all the numbers together, and all the variables

-1Y^2+5Y+6=0

a = -1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·(-1)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-1}=\frac{-12}{-2}
=+6
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-1}=\frac{2}{-2}
=-1
$