(x+1)+(x+-3)=(6x2+-6)

Solution for (x+1)+(x+-3)=(6x2+-6) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

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

We get rid of parentheses

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


We calculate terms in parentheses: -((6x^2-6)), so:

(6x^2-6)

We get rid of parentheses

6x^2-6

Back to the equation:

-(6x^2-6)


We add all the numbers together, and all the variables

2x-(6x^2-6)-2=0

We get rid of parentheses

-6x^2+2x+6-2=0

We add all the numbers together, and all the variables

-6x^2+2x+4=0

a = -6; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-6)·4
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-6}=\frac{-12}{-12}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-6}=\frac{8}{-12}
=-2/3
$