(3y-5)(1-y)=0

Solution for (3y-5)(1-y)=0 equation:

(3y-5)(1-y)=0

We add all the numbers together, and all the variables

(3y-5)(-1y+1)=0

We multiply parentheses ..

(-3y^2+3y+5y-5)=0

We get rid of parentheses

-3y^2+3y+5y-5=0

We add all the numbers together, and all the variables

-3y^2+8y-5=0

a = -3; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·(-3)·(-5)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*-3}=\frac{-10}{-6}
=1+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*-3}=\frac{-6}{-6}
=1
$