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

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

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

We multiply parentheses ..

(+16x^2+2x-40x-5)=0

We get rid of parentheses

16x^2+2x-40x-5=0

We add all the numbers together, and all the variables

16x^2-38x-5=0

a = 16; b = -38; c = -5;
Δ = b2-4ac
Δ = -382-4·16·(-5)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-42}{2*16}=\frac{-4}{32}
=-1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+42}{2*16}=\frac{80}{32}
=2+1/2
$