8(x^2-2x+4)=2x(x+1)-8

Solution for 8(x^2-2x+4)=2x(x+1)-8 equation:

8(x^2-2x+4)=2x(x+1)-8

We move all terms to the left:

8(x^2-2x+4)-(2x(x+1)-8)=0

We multiply parentheses

8x^2-16x-(2x(x+1)-8)+32=0


We calculate terms in parentheses: -(2x(x+1)-8), so:

2x(x+1)-8

We multiply parentheses

2x^2+2x-8

Back to the equation:

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


We get rid of parentheses

8x^2-2x^2-16x-2x+8+32=0

We add all the numbers together, and all the variables

6x^2-18x+40=0

a = 6; b = -18; c = +40;
Δ = b2-4ac
Δ = -182-4·6·40
Δ = -636
Delta is less than zero, so there is no solution for the equation