(2y-4)-4y(6y-10)=0

Solution for (2y-4)-4y(6y-10)=0 equation:

(2y-4)-4y(6y-10)=0

We multiply parentheses

-24y^2+(2y-4)+40y=0

We get rid of parentheses

-24y^2+2y+40y-4=0

We add all the numbers together, and all the variables

-24y^2+42y-4=0

a = -24; b = 42; c = -4;
Δ = b2-4ac
Δ = 422-4·(-24)·(-4)
Δ = 1380
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1380}=\sqrt{4*345}=\sqrt{4}*\sqrt{345}=2\sqrt{345}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{345}}{2*-24}=\frac{-42-2\sqrt{345}}{-48}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{345}}{2*-24}=\frac{-42+2\sqrt{345}}{-48}
$