(2x+1)(5x-8)=0

Solution for (2x+1)(5x-8)=0 equation:

(2x+1)(5x-8)=0

We multiply parentheses ..

(+10x^2-16x+5x-8)=0

We get rid of parentheses

10x^2-16x+5x-8=0

We add all the numbers together, and all the variables

10x^2-11x-8=0

a = 10; b = -11; c = -8;
Δ = b2-4ac
Δ = -112-4·10·(-8)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-21}{2*10}=\frac{-10}{20}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+21}{2*10}=\frac{32}{20}
=1+3/5
$