(8x-3)(-2x+1)=0

Solution for (8x-3)(-2x+1)=0 equation:

(8x-3)(-2x+1)=0

We multiply parentheses ..

(-16x^2+8x+6x-3)=0

We get rid of parentheses

-16x^2+8x+6x-3=0

We add all the numbers together, and all the variables

-16x^2+14x-3=0

a = -16; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·(-16)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*-16}=\frac{-16}{-32}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*-16}=\frac{-12}{-32}
=3/8
$