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

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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 $

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