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

Solution for 6y^2+12y+3=y^2-1 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5y^2+12y+4=0

a = 5; b = 12; c = +4;
Δ = b2-4ac
Δ = 122-4·5·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*5}=\frac{-20}{10}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*5}=\frac{-4}{10}
=-2/5
$