540=x^2+(0.6x)

Solution for 540=x^2+(0.6x) equation:

540=x^2+(0.6x)

We move all terms to the left:

540-(x^2+(0.6x))=0

We add all the numbers together, and all the variables

-(x^2+(+0.6x))+540=0


We calculate terms in parentheses: -(x^2+(+0.6x)), so:

x^2+(+0.6x)

We get rid of parentheses

x^2+0.6x

Back to the equation:

-(x^2+0.6x)


We get rid of parentheses

-x^2-0.6x+540=0

We add all the numbers together, and all the variables

-1x^2-0.6x+540=0

a = -1; b = -0.6; c = +540;
Δ = b2-4ac
Δ = -0.62-4·(-1)·540
Δ = 2160.36
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{-(-0.6)-\sqrt{2160.36}}{2*-1}=\frac{0.6-\sqrt{2160.36}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.6)+\sqrt{2160.36}}{2*-1}=\frac{0.6+\sqrt{2160.36}}{-2}
$