(6x-5)(8x+1)=0

Solution for (6x-5)(8x+1)=0 equation:

(6x-5)(8x+1)=0

We multiply parentheses ..

(+48x^2+6x-40x-5)=0

We get rid of parentheses

48x^2+6x-40x-5=0

We add all the numbers together, and all the variables

48x^2-34x-5=0

a = 48; b = -34; c = -5;
Δ = b2-4ac
Δ = -342-4·48·(-5)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-46}{2*48}=\frac{-12}{96}
=-1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+46}{2*48}=\frac{80}{96}
=5/6
$