40=x^2/20+x

Solution for 40=x^2/20+x equation:

40=x^2/20+x

We move all terms to the left:

40-(x^2/20+x)=0


Domain of the equation: 20+x)!=0

We move all terms containing x to the left, all other terms to the right

x)!=-20

x!=-20/1

x!=-20

x∈R


We get rid of parentheses

-x^2/20-x+40=0

We multiply all the terms by the denominator


-x^2-x*20+40*20=0

We add all the numbers together, and all the variables

-1x^2-x*20+800=0

Wy multiply elements

-1x^2-20x+800=0

a = -1; b = -20; c = +800;
Δ = b2-4ac
Δ = -202-4·(-1)·800
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-60}{2*-1}=\frac{-40}{-2}
=+20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+60}{2*-1}=\frac{80}{-2}
=-40
$