3x^2+6x+x^2-2x-1=x+1

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

3x^2+6x+x^2-2x-1=x+1

We move all terms to the left:

3x^2+6x+x^2-2x-1-(x+1)=0

We add all the numbers together, and all the variables

4x^2+4x-(x+1)-1=0

We get rid of parentheses

4x^2+4x-x-1-1=0

We add all the numbers together, and all the variables

4x^2+3x-2=0

a = 4; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·4·(-2)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{41}}{2*4}=\frac{-3-\sqrt{41}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*4}=\frac{-3+\sqrt{41}}{8}
$