62=x+2x+(2+x^2)

Solution for 62=x+2x+(2+x^2) equation:

62=x+2x+(2+x^2)

We move all terms to the left:

62-(x+2x+(2+x^2))=0


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

x+2x+(2+x^2)

determiningTheFunctionDomain
(2+x^2)+x+2x

We add all the numbers together, and all the variables

(2+x^2)+3x

We get rid of parentheses

x^2+3x+2

Back to the equation:

-(x^2+3x+2)


We get rid of parentheses

-x^2-3x-2+62=0

We add all the numbers together, and all the variables

-1x^2-3x+60=0

a = -1; b = -3; c = +60;
Δ = b2-4ac
Δ = -32-4·(-1)·60
Δ = 249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{249}}{2*-1}=\frac{3-\sqrt{249}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{249}}{2*-1}=\frac{3+\sqrt{249}}{-2}
$