(8x-3)(x-4)=0

Solution for (8x-3)(x-4)=0 equation:

(8x-3)(x-4)=0

We multiply parentheses ..

(+8x^2-32x-3x+12)=0

We get rid of parentheses

8x^2-32x-3x+12=0

We add all the numbers together, and all the variables

8x^2-35x+12=0

a = 8; b = -35; c = +12;
Δ = b2-4ac
Δ = -352-4·8·12
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{-(-35)-29}{2*8}=\frac{6}{16}
=3/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+29}{2*8}=\frac{64}{16}
=4
$