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

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

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

We multiply parentheses ..

(+10x^2+16x-25x-40)=0

We get rid of parentheses

10x^2+16x-25x-40=0

We add all the numbers together, and all the variables

10x^2-9x-40=0

a = 10; b = -9; c = -40;
Δ = b2-4ac
Δ = -92-4·10·(-40)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-41}{2*10}=\frac{-32}{20}
=-1+3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+41}{2*10}=\frac{50}{20}
=2+1/2
$