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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

4x-2x-((8x+6)(x-1)+22)+6=0

We multiply parentheses ..

-((+8x^2-8x+6x-6)+22)+4x-2x+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^2-2x+16

Back to the equation:

-(8x^2-2x+16)



We add all the numbers together, and all the variables

2x-(8x^2-2x+16)+6=0

We get rid of parentheses

-8x^2+2x+2x-16+6=0

We add all the numbers together, and all the variables

-8x^2+4x-10=0

a = -8; b = 4; c = -10;
Δ = b2-4ac
Δ = 42-4·(-8)·(-10)
Δ = -304
Delta is less than zero, so there is no solution for the equation

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