(2x-4)(8x+3)=0

Solution for (2x-4)(8x+3)=0 equation:

(2x-4)(8x+3)=0

We multiply parentheses ..

(+16x^2+6x-32x-12)=0

We get rid of parentheses

16x^2+6x-32x-12=0

We add all the numbers together, and all the variables

16x^2-26x-12=0

a = 16; b = -26; c = -12;
Δ = b2-4ac
Δ = -262-4·16·(-12)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-38}{2*16}=\frac{-12}{32}
=-3/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+38}{2*16}=\frac{64}{32}
=2
$