0=(125+5x)(100-2x)-(500+30(100-2x))

Solution for 0=(125+5x)(100-2x)-(500+30(100-2x)) equation:

0=(125+5x)(100-2x)-(500+30(100-2x))

We move all terms to the left:

0-((125+5x)(100-2x)-(500+30(100-2x)))=0

We add all the numbers together, and all the variables

-((5x+125)(-2x+100)-(500+30(-2x+100)))+0=0

We add all the numbers together, and all the variables

-((5x+125)(-2x+100)-(500+30(-2x+100)))=0

We multiply parentheses ..

-((-10x^2+500x-250x+12500)-(500+30(-2x+100)))=0


We calculate terms in parentheses: -((-10x^2+500x-250x+12500)-(500+30(-2x+100))), so:

(-10x^2+500x-250x+12500)-(500+30(-2x+100))

We get rid of parentheses

-10x^2+500x-250x-(500+30(-2x+100))+12500


We calculate terms in parentheses: -(500+30(-2x+100)), so:

500+30(-2x+100)

determiningTheFunctionDomain
30(-2x+100)+500

We multiply parentheses

-60x+3000+500

We add all the numbers together, and all the variables

-60x+3500

Back to the equation:

-(-60x+3500)


We add all the numbers together, and all the variables

-10x^2+250x-(-60x+3500)+12500

We get rid of parentheses

-10x^2+250x+60x-3500+12500

We add all the numbers together, and all the variables

-10x^2+310x+9000

Back to the equation:

-(-10x^2+310x+9000)


We get rid of parentheses

10x^2-310x-9000=0

a = 10; b = -310; c = -9000;
Δ = b2-4ac
Δ = -3102-4·10·(-9000)
Δ = 456100
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:

$\sqrt{\Delta}=\sqrt{456100}=\sqrt{100*4561}=\sqrt{100}*\sqrt{4561}=10\sqrt{4561}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-310)-10\sqrt{4561}}{2*10}=\frac{310-10\sqrt{4561}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-310)+10\sqrt{4561}}{2*10}=\frac{310+10\sqrt{4561}}{20}
$