x*(2x-5)+8x=x(3+x)

Solution for x*(2x-5)+8x=x(3+x) equation:

x(2x-5)+8x=x(3+x)

We move all terms to the left:

x(2x-5)+8x-(x(3+x))=0

We add all the numbers together, and all the variables

x(2x-5)+8x-(x(x+3))=0

We add all the numbers together, and all the variables

8x+x(2x-5)-(x(x+3))=0

We multiply parentheses

2x^2+8x-5x-(x(x+3))=0


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

x(x+3)

We multiply parentheses

x^2+3x

Back to the equation:

-(x^2+3x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-x^2+3x-3x=0

We add all the numbers together, and all the variables

x^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

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