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

Solution for (2-y)(4y-3)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4y^2+3y+8y-6)=0

We get rid of parentheses

-4y^2+3y+8y-6=0

We add all the numbers together, and all the variables

-4y^2+11y-6=0

a = -4; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·(-4)·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*-4}=\frac{-16}{-8}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*-4}=\frac{-6}{-8}
=3/4
$