3x^2-24x+x-8=0

Solution for 3x^2-24x+x-8=0 equation:

3x^2-24x+x-8=0

We add all the numbers together, and all the variables

3x^2-23x-8=0

a = 3; b = -23; c = -8;
Δ = b2-4ac
Δ = -232-4·3·(-8)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-25}{2*3}=\frac{-2}{6}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+25}{2*3}=\frac{48}{6}
=8
$