360=10x+x^2+(5x+36)

Solution for 360=10x+x^2+(5x+36) equation:

360=10x+x^2+(5x+36)

We move all terms to the left:

360-(10x+x^2+(5x+36))=0


We calculate terms in parentheses: -(10x+x^2+(5x+36)), so:

10x+x^2+(5x+36)

determiningTheFunctionDomain
x^2+10x+(5x+36)

We get rid of parentheses

x^2+10x+5x+36

We add all the numbers together, and all the variables

x^2+15x+36

Back to the equation:

-(x^2+15x+36)


We get rid of parentheses

-x^2-15x-36+360=0

We add all the numbers together, and all the variables

-1x^2-15x+324=0

a = -1; b = -15; c = +324;
Δ = b2-4ac
Δ = -152-4·(-1)·324
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-39}{2*-1}=\frac{-24}{-2}
=+12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+39}{2*-1}=\frac{54}{-2}
=-27
$