(32x+x)(x)=(3x)(3x+8)

Solution for (32x+x)(x)=(3x)(3x+8) equation:

(32x+x)(x)=(3x)(3x+8)

We move all terms to the left:

(32x+x)(x)-((3x)(3x+8))=0

We add all the numbers together, and all the variables

(+33x)x-(3x(3x+8))=0

We multiply parentheses

33x^2-(3x(3x+8))=0


We calculate terms in parentheses: -(3x(3x+8)), so:

3x(3x+8)

We multiply parentheses

9x^2+24x

Back to the equation:

-(9x^2+24x)


We get rid of parentheses

33x^2-9x^2-24x=0

We add all the numbers together, and all the variables

24x^2-24x=0

a = 24; b = -24; c = 0;
Δ = b2-4ac
Δ = -242-4·24·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24}{2*24}=\frac{0}{48}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24}{2*24}=\frac{48}{48}
=1
$