(3x-13)(2x+4)+x=180

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Solution for (3x-13)(2x+4)+x=180 equation:



(3x-13)(2x+4)+x=180
We move all terms to the left:
(3x-13)(2x+4)+x-(180)=0
We add all the numbers together, and all the variables
x+(3x-13)(2x+4)-180=0
We multiply parentheses ..
(+6x^2+12x-26x-52)+x-180=0
We get rid of parentheses
6x^2+12x-26x+x-52-180=0
We add all the numbers together, and all the variables
6x^2-13x-232=0
a = 6; b = -13; c = -232;
Δ = b2-4ac
Δ = -132-4·6·(-232)
Δ = 5737
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{5737}}{2*6}=\frac{13-\sqrt{5737}}{12} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{5737}}{2*6}=\frac{13+\sqrt{5737}}{12} $

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