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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(-4Y+8)-(Y(Y+3)-4)=0


We calculate terms in parentheses: -(Y(Y+3)-4), so:

Y(Y+3)-4

We multiply parentheses

Y^2+3Y-4

Back to the equation:

-(Y^2+3Y-4)


We get rid of parentheses

-Y^2-4Y-3Y+8+4=0

We add all the numbers together, and all the variables

-1Y^2-7Y+12=0

a = -1; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·(-1)·12
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{97}}{2*-1}=\frac{7-\sqrt{97}}{-2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{97}}{2*-1}=\frac{7+\sqrt{97}}{-2}
$