(3y-8)(4y-1)=0

Solution for (3y-8)(4y-1)=0 equation:

(3y-8)(4y-1)=0

We multiply parentheses ..

(+12y^2-3y-32y+8)=0

We get rid of parentheses

12y^2-3y-32y+8=0

We add all the numbers together, and all the variables

12y^2-35y+8=0

a = 12; b = -35; c = +8;
Δ = b2-4ac
Δ = -352-4·12·8
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-29}{2*12}=\frac{6}{24}
=1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+29}{2*12}=\frac{64}{24}
=2+2/3
$