(6-5x)(4x+1)=0

Solution for (6-5x)(4x+1)=0 equation:

(6-5x)(4x+1)=0

We add all the numbers together, and all the variables

(-5x+6)(4x+1)=0

We multiply parentheses ..

(-20x^2-5x+24x+6)=0

We get rid of parentheses

-20x^2-5x+24x+6=0

We add all the numbers together, and all the variables

-20x^2+19x+6=0

a = -20; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·(-20)·6
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-29}{2*-20}=\frac{-48}{-40}
=1+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+29}{2*-20}=\frac{10}{-40}
=-1/4
$