(3-y)(y-3)=-4

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

(3-y)(y-3)=-4

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1y^2+3y+3y-9)+4=0

We get rid of parentheses

-1y^2+3y+3y-9+4=0

We add all the numbers together, and all the variables

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

a = -1; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·(-1)·(-5)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*-1}=\frac{-10}{-2}
=+5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*-1}=\frac{-2}{-2}
=1
$