(4y-4)(7y-5)=0

Solution for (4y-4)(7y-5)=0 equation:

(4y-4)(7y-5)=0

We multiply parentheses ..

(+28y^2-20y-28y+20)=0

We get rid of parentheses

28y^2-20y-28y+20=0

We add all the numbers together, and all the variables

28y^2-48y+20=0

a = 28; b = -48; c = +20;
Δ = b2-4ac
Δ = -482-4·28·20
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8}{2*28}=\frac{40}{56}
=5/7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8}{2*28}=\frac{56}{56}
=1
$