(2x+3)2+(2x-3)2=(8x+6)(X+1)+22

Solution for (2x+3)2+(2x-3)2=(8x+6)(X+1)+22 equation:

(2x+3)2+(2x-3)2=(8x+6)(x+1)+22

We move all terms to the left:

(2x+3)2+(2x-3)2-((8x+6)(x+1)+22)=0

We multiply parentheses

4x+4x-((8x+6)(x+1)+22)+6-6=0

We multiply parentheses ..

-((+8x^2+8x+6x+6)+22)+4x+4x+6-6=0


We calculate terms in parentheses: -((+8x^2+8x+6x+6)+22), so:

(+8x^2+8x+6x+6)+22

We get rid of parentheses

8x^2+8x+6x+6+22

We add all the numbers together, and all the variables

8x^2+14x+28

Back to the equation:

-(8x^2+14x+28)


We add all the numbers together, and all the variables

8x-(8x^2+14x+28)=0

We get rid of parentheses

-8x^2+8x-14x-28=0

We add all the numbers together, and all the variables

-8x^2-6x-28=0

a = -8; b = -6; c = -28;
Δ = b2-4ac
Δ = -62-4·(-8)·(-28)
Δ = -860
Delta is less than zero, so there is no solution for the equation