(40-2x)=0,6x(x+5)

Solution for (40-2x)=0,6x(x+5) equation:

(40-2x)=0.6x(x+5)

We move all terms to the left:

(40-2x)-(0.6x(x+5))=0

We add all the numbers together, and all the variables

(-2x+40)-(0.6x(x+5))=0

We get rid of parentheses

-2x-(0.6x(x+5))+40=0


We calculate terms in parentheses: -(0.6x(x+5)), so:

0.6x(x+5)

We multiply parentheses

0x^2+0x

We add all the numbers together, and all the variables

x^2+x

Back to the equation:

-(x^2+x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-3x+40=0

a = -1; b = -3; c = +40;
Δ = b2-4ac
Δ = -32-4·(-1)·40
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*-1}=\frac{16}{-2}
=-8
$