(32-x^2)=(x^2+20x+80)

Solution for (32-x^2)=(x^2+20x+80) equation:

(32-x^2)=(x^2+20x+80)

We move all terms to the left:

(32-x^2)-((x^2+20x+80))=0

We get rid of parentheses

-x^2-((x^2+20x+80))+32=0


We calculate terms in parentheses: -((x^2+20x+80)), so:

(x^2+20x+80)

We get rid of parentheses

x^2+20x+80

Back to the equation:

-(x^2+20x+80)


We add all the numbers together, and all the variables

-1x^2-(x^2+20x+80)+32=0

We get rid of parentheses

-1x^2-x^2-20x-80+32=0

We add all the numbers together, and all the variables

-2x^2-20x-48=0

a = -2; b = -20; c = -48;
Δ = b2-4ac
Δ = -202-4·(-2)·(-48)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*-2}=\frac{16}{-4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*-2}=\frac{24}{-4}
=-6
$