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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+4x^2-10x+2x-5))=0


We calculate terms in parentheses: -((+4x^2-10x+2x-5)), so:

(+4x^2-10x+2x-5)

We get rid of parentheses

4x^2-10x+2x-5

We add all the numbers together, and all the variables

4x^2-8x-5

Back to the equation:

-(4x^2-8x-5)


We get rid of parentheses

-4x^2+8x+5=0

a = -4; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·(-4)·5
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*-4}=\frac{-20}{-8}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*-4}=\frac{4}{-8}
=-1/2
$