(2x+30)(3x-20)=180

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Solution for (2x+30)(3x-20)=180 equation:



(2x+30)(3x-20)=180
We move all terms to the left:
(2x+30)(3x-20)-(180)=0
We multiply parentheses ..
(+6x^2-40x+90x-600)-180=0
We get rid of parentheses
6x^2-40x+90x-600-180=0
We add all the numbers together, and all the variables
6x^2+50x-780=0
a = 6; b = 50; c = -780;
Δ = b2-4ac
Δ = 502-4·6·(-780)
Δ = 21220
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{21220}=\sqrt{4*5305}=\sqrt{4}*\sqrt{5305}=2\sqrt{5305}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{5305}}{2*6}=\frac{-50-2\sqrt{5305}}{12} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{5305}}{2*6}=\frac{-50+2\sqrt{5305}}{12} $

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