4(3+7y)+6(2-y)y=2

Solution for 4(3+7y)+6(2-y)y=2 equation:

4(3+7y)+6(2-y)y=2

We move all terms to the left:

4(3+7y)+6(2-y)y-(2)=0

We add all the numbers together, and all the variables

4(7y+3)+6(-1y+2)y-2=0

We multiply parentheses

-6y^2+28y+12y+12-2=0

We add all the numbers together, and all the variables

-6y^2+40y+10=0

a = -6; b = 40; c = +10;
Δ = b2-4ac
Δ = 402-4·(-6)·10
Δ = 1840
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:

$\sqrt{\Delta}=\sqrt{1840}=\sqrt{16*115}=\sqrt{16}*\sqrt{115}=4\sqrt{115}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{115}}{2*-6}=\frac{-40-4\sqrt{115}}{-12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{115}}{2*-6}=\frac{-40+4\sqrt{115}}{-12}
$