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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+3x^2-8x+9x-24))=0


We calculate terms in parentheses: -((+3x^2-8x+9x-24)), so:

(+3x^2-8x+9x-24)

We get rid of parentheses

3x^2-8x+9x-24

We add all the numbers together, and all the variables

3x^2+x-24

Back to the equation:

-(3x^2+x-24)


We get rid of parentheses

-3x^2-x+24=0

We add all the numbers together, and all the variables

-3x^2-1x+24=0

a = -3; b = -1; c = +24;
Δ = b2-4ac
Δ = -12-4·(-3)·24
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*-3}=\frac{-16}{-6}
=2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*-3}=\frac{18}{-6}
=-3
$