2x(3x+4)=9x+12

Solution for 2x(3x+4)=9x+12 equation:

2x(3x+4)=9x+12

We move all terms to the left:

2x(3x+4)-(9x+12)=0

We multiply parentheses

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

We get rid of parentheses

6x^2+8x-9x-12=0

We add all the numbers together, and all the variables

6x^2-1x-12=0

a = 6; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·6·(-12)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*6}=\frac{-16}{12}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*6}=\frac{18}{12}
=1+1/2
$