6+y(12y+3)=15-y

Solution for 6+y(12y+3)=15-y equation:

6+y(12y+3)=15-y

We move all terms to the left:

6+y(12y+3)-(15-y)=0

We add all the numbers together, and all the variables

y(12y+3)-(-1y+15)+6=0

We multiply parentheses

12y^2+3y-(-1y+15)+6=0

We get rid of parentheses

12y^2+3y+1y-15+6=0

We add all the numbers together, and all the variables

12y^2+4y-9=0

a = 12; b = 4; c = -9;
Δ = b2-4ac
Δ = 42-4·12·(-9)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{7}}{2*12}=\frac{-4-8\sqrt{7}}{24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{7}}{2*12}=\frac{-4+8\sqrt{7}}{24}
$