x=(x+140)(3x-20)

Solution for x=(x+140)(3x-20) equation:

x=(x+140)(3x-20)

We move all terms to the left:

x-((x+140)(3x-20))=0

We multiply parentheses ..

-((+3x^2-20x+420x-2800))+x=0


We calculate terms in parentheses: -((+3x^2-20x+420x-2800)), so:

(+3x^2-20x+420x-2800)

We get rid of parentheses

3x^2-20x+420x-2800

We add all the numbers together, and all the variables

3x^2+400x-2800

Back to the equation:

-(3x^2+400x-2800)


We add all the numbers together, and all the variables

x-(3x^2+400x-2800)=0

We get rid of parentheses

-3x^2+x-400x+2800=0

We add all the numbers together, and all the variables

-3x^2-399x+2800=0

a = -3; b = -399; c = +2800;
Δ = b2-4ac
Δ = -3992-4·(-3)·2800
Δ = 192801
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{-(-399)-\sqrt{192801}}{2*-3}=\frac{399-\sqrt{192801}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-399)+\sqrt{192801}}{2*-3}=\frac{399+\sqrt{192801}}{-6}
$