(2x+1)(10x-5)=0

Solution for (2x+1)(10x-5)=0 equation:

(2x+1)(10x-5)=0

We multiply parentheses ..

(+20x^2-10x+10x-5)=0

We get rid of parentheses

20x^2-10x+10x-5=0

We add all the numbers together, and all the variables

20x^2-5=0

a = 20; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·20·(-5)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*20}=\frac{-20}{40}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*20}=\frac{20}{40}
=1/2
$