((-32x^2)/3600)+x+180=100

Solution for ((-32x^2)/3600)+x+180=100 equation:

((-32x^2)/3600)+x+180=100

We move all terms to the left:

((-32x^2)/3600)+x+180-(100)=0

We add all the numbers together, and all the variables

((-32x^2)/3600)+x+80=0

We multiply all the terms by the denominator


((-32x^2)+x*3600)+80*3600)=0


We calculate terms in parentheses: +((-32x^2)+x*3600), so:

(-32x^2)+x*3600

Wy multiply elements

(-32x^2)+3600x

We get rid of parentheses

-32x^2+3600x

Back to the equation:

+(-32x^2+3600x)


We add all the numbers together, and all the variables

(-32x^2+3600x)=0

We get rid of parentheses

-32x^2+3600x=0

a = -32; b = 3600; c = 0;
Δ = b2-4ac
Δ = 36002-4·(-32)·0
Δ = 12960000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12960000}=3600$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3600)-3600}{2*-32}=\frac{-7200}{-64}
=112+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3600)+3600}{2*-32}=\frac{0}{-64}
=0
$