0=(x+2)(3x-17)

Solution for 0=(x+2)(3x-17) equation:

0=(x+2)(3x-17)

We move all terms to the left:

0-((x+2)(3x-17))=0

We add all the numbers together, and all the variables

-((x+2)(3x-17))=0

We multiply parentheses ..

-((+3x^2-17x+6x-34))=0


We calculate terms in parentheses: -((+3x^2-17x+6x-34)), so:

(+3x^2-17x+6x-34)

We get rid of parentheses

3x^2-17x+6x-34

We add all the numbers together, and all the variables

3x^2-11x-34

Back to the equation:

-(3x^2-11x-34)


We get rid of parentheses

-3x^2+11x+34=0

a = -3; b = 11; c = +34;
Δ = b2-4ac
Δ = 112-4·(-3)·34
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-23}{2*-3}=\frac{-34}{-6}
=5+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+23}{2*-3}=\frac{12}{-6}
=-2
$