2x(3x-7)+4x(3x+2)=6x(5x+9)

Solution for 2x(3x-7)+4x(3x+2)=6x(5x+9) equation:

2x(3x-7)+4x(3x+2)=6x(5x+9)

We move all terms to the left:

2x(3x-7)+4x(3x+2)-(6x(5x+9))=0

We multiply parentheses

6x^2+12x^2-14x+8x-(6x(5x+9))=0


We calculate terms in parentheses: -(6x(5x+9)), so:

6x(5x+9)

We multiply parentheses

30x^2+54x

Back to the equation:

-(30x^2+54x)


We add all the numbers together, and all the variables

18x^2-6x-(30x^2+54x)=0

We get rid of parentheses

18x^2-30x^2-6x-54x=0

We add all the numbers together, and all the variables

-12x^2-60x=0

a = -12; b = -60; c = 0;
Δ = b2-4ac
Δ = -602-4·(-12)·0
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-60}{2*-12}=\frac{0}{-24}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+60}{2*-12}=\frac{120}{-24}
=-5
$