(7x-4)(8x+1)=0

Solution for (7x-4)(8x+1)=0 equation:

(7x-4)(8x+1)=0

We multiply parentheses ..

(+56x^2+7x-32x-4)=0

We get rid of parentheses

56x^2+7x-32x-4=0

We add all the numbers together, and all the variables

56x^2-25x-4=0

a = 56; b = -25; c = -4;
Δ = b2-4ac
Δ = -252-4·56·(-4)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-39}{2*56}=\frac{-14}{112}
=-1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+39}{2*56}=\frac{64}{112}
=4/7
$