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

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

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

We multiply parentheses ..

(+2x^2-3x-16x+24)=0

We get rid of parentheses

2x^2-3x-16x+24=0

We add all the numbers together, and all the variables

2x^2-19x+24=0

a = 2; b = -19; c = +24;
Δ = b2-4ac
Δ = -192-4·2·24
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*2}=\frac{6}{4}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*2}=\frac{32}{4}
=8
$