0=(1200-100x)(10+2x)

Solution for 0=(1200-100x)(10+2x) equation:

0=(1200-100x)(10+2x)

We move all terms to the left:

0-((1200-100x)(10+2x))=0

We add all the numbers together, and all the variables

-((-100x+1200)(2x+10))+0=0

We add all the numbers together, and all the variables

-((-100x+1200)(2x+10))=0

We multiply parentheses ..

-((-200x^2-1000x+2400x+12000))=0


We calculate terms in parentheses: -((-200x^2-1000x+2400x+12000)), so:

(-200x^2-1000x+2400x+12000)

We get rid of parentheses

-200x^2-1000x+2400x+12000

We add all the numbers together, and all the variables

-200x^2+1400x+12000

Back to the equation:

-(-200x^2+1400x+12000)


We get rid of parentheses

200x^2-1400x-12000=0

a = 200; b = -1400; c = -12000;
Δ = b2-4ac
Δ = -14002-4·200·(-12000)
Δ = 11560000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11560000}=3400$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1400)-3400}{2*200}=\frac{-2000}{400}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1400)+3400}{2*200}=\frac{4800}{400}
=12
$