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

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

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

We add all the numbers together, and all the variables

(2x+3)(3x+4)=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+17x+12=0

a = 6; b = 17; c = +12;
Δ = b2-4ac
Δ = 172-4·6·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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