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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(2(x^2+3)+1), so:

2(x^2+3)+1

We multiply parentheses

2x^2+6+1

We add all the numbers together, and all the variables

2x^2+7

Back to the equation:

-(2x^2+7)


We get rid of parentheses

-2x^2+(2x+1)^2-7=0

We move all terms containing x to the left, all other terms to the right

-2x^2+(2x+1)^2=7