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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:We add all the numbers together, and all the variables
(+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+440x+2800
Back to the equation:
-(3x^2+440x+2800)
x-(3x^2+440x+2800)=0
We get rid of parentheses
-3x^2+x-440x-2800=0
We add all the numbers together, and all the variables
-3x^2-439x-2800=0
a = -3; b = -439; c = -2800;
Δ = b2-4ac
Δ = -4392-4·(-3)·(-2800)
Δ = 159121
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{-(-439)-\sqrt{159121}}{2*-3}=\frac{439-\sqrt{159121}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-439)+\sqrt{159121}}{2*-3}=\frac{439+\sqrt{159121}}{-6} $
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