(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

(y+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
$