8x=4(x2+2x)

Solution for 8x=4(x2+2x) equation:

8x=4(x2+2x)

We move all terms to the left:

8x-(4(x2+2x))=0

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(4(+x^2+2x)), so:

4(+x^2+2x)

We multiply parentheses

4x^2+8x

Back to the equation:

-(4x^2+8x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2=0

a = -4; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-4)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{-8}=0$