0=(9x-3)(2x+4)

Solution for 0=(9x-3)(2x+4) equation:

0=(9x-3)(2x+4)

We move all terms to the left:

0-((9x-3)(2x+4))=0

We add all the numbers together, and all the variables

-((9x-3)(2x+4))=0

We multiply parentheses ..

-((+18x^2+36x-6x-12))=0


We calculate terms in parentheses: -((+18x^2+36x-6x-12)), so:

(+18x^2+36x-6x-12)

We get rid of parentheses

18x^2+36x-6x-12

We add all the numbers together, and all the variables

18x^2+30x-12

Back to the equation:

-(18x^2+30x-12)


We get rid of parentheses

-18x^2-30x+12=0

a = -18; b = -30; c = +12;
Δ = b2-4ac
Δ = -302-4·(-18)·12
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-42}{2*-18}=\frac{-12}{-36}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+42}{2*-18}=\frac{72}{-36}
=-2
$