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

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

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

We multiply parentheses ..

(+8x^2-16x+6x-12)=0

We get rid of parentheses

8x^2-16x+6x-12=0

We add all the numbers together, and all the variables

8x^2-10x-12=0

a = 8; b = -10; c = -12;
Δ = b2-4ac
Δ = -102-4·8·(-12)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*8}=\frac{-12}{16}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*8}=\frac{32}{16}
=2
$