2500=(480-8x)(5+.1x)

Solution for 2500=(480-8x)(5+.1x) equation:

2500=(480-8x)(5+.1x)

We move all terms to the left:

2500-((480-8x)(5+.1x))=0

We add all the numbers together, and all the variables

-((-8x+480)(.1x+5))+2500=0

We multiply parentheses ..

-((-8x^2-40x+480x+2400))+2500=0


We calculate terms in parentheses: -((-8x^2-40x+480x+2400)), so:

(-8x^2-40x+480x+2400)

We get rid of parentheses

-8x^2-40x+480x+2400

We add all the numbers together, and all the variables

-8x^2+440x+2400

Back to the equation:

-(-8x^2+440x+2400)


We get rid of parentheses

8x^2-440x-2400+2500=0

We add all the numbers together, and all the variables

8x^2-440x+100=0

a = 8; b = -440; c = +100;
Δ = b2-4ac
Δ = -4402-4·8·100
Δ = 190400
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:

$\sqrt{\Delta}=\sqrt{190400}=\sqrt{1600*119}=\sqrt{1600}*\sqrt{119}=40\sqrt{119}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-440)-40\sqrt{119}}{2*8}=\frac{440-40\sqrt{119}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-440)+40\sqrt{119}}{2*8}=\frac{440+40\sqrt{119}}{16}
$