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

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

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

We add all the numbers together, and all the variables

(-1y+6)(4y+2)=0

We multiply parentheses ..

(-4y^2-2y+24y+12)=0

We get rid of parentheses

-4y^2-2y+24y+12=0

We add all the numbers together, and all the variables

-4y^2+22y+12=0

a = -4; b = 22; c = +12;
Δ = b2-4ac
Δ = 222-4·(-4)·12
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*-4}=\frac{-48}{-8}
=+6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*-4}=\frac{4}{-8}
=-1/2
$