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

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

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

We multiply parentheses ..

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

We get rid of parentheses

10x^2-2x+5x-1=0

We add all the numbers together, and all the variables

10x^2+3x-1=0

a = 10; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·10·(-1)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*10}=\frac{-10}{20}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*10}=\frac{4}{20}
=1/5
$