(2x-4)(5x+3)=0

Solution for (2x-4)(5x+3)=0 equation:

(2x-4)(5x+3)=0

We multiply parentheses ..

(+10x^2+6x-20x-12)=0

We get rid of parentheses

10x^2+6x-20x-12=0

We add all the numbers together, and all the variables

10x^2-14x-12=0

a = 10; b = -14; c = -12;
Δ = b2-4ac
Δ = -142-4·10·(-12)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-26}{2*10}=\frac{-12}{20}
=-3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*10}=\frac{40}{20}
=2
$