x^+x(x-6)=0)x(x+3)=5x

Solution for x^+x(x-6)=0)x(x+3)=5x equation:

x^+x(x-6)=0)x(x+3)=5x

We move all terms to the left:

x^+x(x-6)-(0)x(x+3))=0

We add all the numbers together, and all the variables

x+x(x-6)-0x(x+3))=0

We multiply parentheses

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

We add all the numbers together, and all the variables

-1x^2-5x=0

a = -1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-1)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-1}=\frac{0}{-2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-1}=\frac{10}{-2}
=-5
$