(5x+1)(2x-4)=0

Solution for (5x+1)(2x-4)=0 equation:

(5x+1)(2x-4)=0

We multiply parentheses ..

(+10x^2-20x+2x-4)=0

We get rid of parentheses

10x^2-20x+2x-4=0

We add all the numbers together, and all the variables

10x^2-18x-4=0

a = 10; b = -18; c = -4;
Δ = b2-4ac
Δ = -182-4·10·(-4)
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{-(-18)-22}{2*10}=\frac{-4}{20}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*10}=\frac{40}{20}
=2
$